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HLEHEMENTS 


OF 


rHOMETRY 


BY 


G. M. SEARLE, C.S.P., 


A.M. (HARV.), A.A.S., FORMERLY ASSISTANT PROFESSOR AT THE U. 8. NAVAL 
ACADEMY, AND ASSISTANT AT THE DUDLEY OBSERVATORY, 
AND AT THAT OF HARVARD COLLEGE. 


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WITH. AN “APPENDIX 
CONTAINING PROBLEMS AND ADDITIONAL PROPOSITIONS. 


NEW YORK: 
JOHN WILEY & SONS, 


15 ASTOR PLACE. 


1827; 





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7 1877. 


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So many and so good text-books on geometry have been published, 
that a new one may certainly seem superfluous. It is, therefore, 
with some hesitation that the present work is submitted to’ the 


public. 


Two motives, principally, have induced its preparation. The first 
has been the desire to reduce ‘what is supposed to be self-evident to 
the smallest possible amount and thus to make the science more 


strictly logical, in opposition, perhaps, to the prevailing tendency at 


. the present day, which seems to be to increase the number of axlomis, 


and base geometry, to a considerable extent, on observation and 
experience. 


Such a course may, no doubt, simplify the subject, if it be not 
earried too far for the capacity of the student. To one of great 
mathematical ability, a very large part, perhaps almost all, of what 
is usually considered as elementary geometry, may seem quite un- 
necessary to be proved; and the experience of common life also cer- 
tainly satisfies people in general of many things in it which are 
ordinarily demonstrated. But, nevertheless, the advantage of sim- 
plicity gained by the multiplication of axioms does not seem to be 
worth the sacrifice of logical accuracy; to assume ten statements to 
be self-evident when eight of them can be deduced from the other 
two, can never cease to be a blemish in a scientific system, and espe- 
cially in one which is studied not by any means so much for the 


sake of acquiring information as for that of acquiring a logical habit 
of mind, 


lv PREFACE. 


In this book the opposite plan has been followed, to the extent of 
not introducing any formal axioms at all, if we except that of Art. 
81. A note on this will be found in Appendix C. Those which are 
not peculiar to geometry, (as that the whole is greater than any of its 
parts) have been tacitly assumed; and others, which are usually so 
assumed, among which may be reckoned the principle of the continuity 


of motion, have been made use of in the same way. 


The three ideas of space, position, and direction, have been made 
the geometrical basis, as stated at the outset. How far this treatment 


is satisfactory will be for the reader to judge. 


The second motive for writing the book is connected with the first, 
though more special in its character. It is to treat the theory of 
parallels in a way open to the least objection. There is an unsatis- 
factoriness in all axioms which have been proposed on this subject 
which has led, in the minds of some, to not only theoretical, but even 
to practical doubts on the ordinary conclusions deduced from them, 
and have given rise to a system without such an axiom, called by va- 
rious names, the most appropriate, as it would seem, being that given 
by J. Bolyai, of “absolute geometry.” This, though an interesting 
subject of study, is too abstruse for beginners; and as the common 
sense of mankind will probably never be much affected by its con- 
clusions, the thing most desirable seems to be to place the convictions 
of common sense, if not on an indisputably logical basis, at least on 
one open to as little objection as possible. This it has been the en- 


deavor of the author to do. 


It may be remarked in this connection, that other doubts on the 
bases of geometrical science have lately been raised; but it has not 


seemed worth while, in a work of this character, to discuss them. 


It remains to make a few remarks on other peculiarities of the 


present treatise. 


1. Definitions have not been given till needed, and the thing to be 


PREFACE. Vv 


defined has been shown to be poathlay if such a demonstration was 
necessary, before defining it. A notable example of such premature 
definitions as are sometimes met with is in that of a plane as “a sur- 
face such that if any two of its points be joined by a straight line, 
that line will lie wholly in the surface.” Such a surface might be an 


impossibility. 


2. The theory of proportions has been omitted, and instead of it 
the equality of fractions has been employed, to which the ordinary 
algebraical rules have been applied. Though the rules of proportion 
are sometimes convenient, the subject seems an unnecessary appendage 
to an elementary geometrical course. The terms “ ratio” and “ pro- 
portional” have been used however in their ordinary and obvious 


’ 


sense. 


3. The name for a figure or structure is understood to apply to its 
perimeter, or surface, and not to its area or volume, following the 
custom employed in the higher mathematics. For instance, a circle 
or ellipse is not understood in analysis to be a space, but a curved 


line. 


4, Algebraic methods have been a little more frequently used in 
the latter part of the book than is usually the case. Ideas usually 
not mtroduced before the study of the calculus have also been intro- 
duced, but not, it is thought, in any way that will give any difficulty 
to the student. 


5. The idea of the formation of lines, areas, and volumes by move- 
ment has been generally adhered to, as a preparation for the higher 
mathematics, and for other reasons, as will be evident at the 


outset. 


6. The solution of problems and the proposition of additional 
theorems has been put off till the appendix, not to interfere with the 
regular logical development of the subject. These problems and 


additional theorems have, for the most part, been taken from the 


V1 PREFACE. 


very complete collection of Chauvenet’s Elementary Geometry ; from 


Hallowell’s Geometrical Analysis, and from Benson’s Geometry. 


It will be noticed that a previous course of elementary algebra is 
supposed. To put algebra before geometry seems to be the natural 
order, as algebra is a science of pure, geometry strictly of applied, 
mathematics. The amount of aigebra required, however, can easily 
be explained to, or acquired by, any average student in a short 


time. 


NEW YORK, March 9, 1877, 





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Mutual Relations of Lines and Planes 


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HLEMENTS OF GHOMETRY. 


eM OMRON RG F 


ANGLES, TRIANGLES, AND PLANES. 


TuERE are three ideas which enter into the definitions of all 
geometrical conceptions, but which are not themselves sus- 
ceptible of geometrical definition. These are space, position, 
and direction. Geometry is the science which develops these 
three ideas. 

1. Definition.— Motion is a continuous change of position in 
space. It is not merely a change of position, it is a continuous 
change ; that is, one in which the geometrical quantities affected 
by the motion change continuously in value, or, in other words, 
in which they pass from one value to another through every 
intermediate value. The admission of this principle of con- 
tinuity in motion may, perhaps, give some difficulty in the 
abstract; but it will hardly admit of doubt in the cases we shall 
have to consider. 

2. Def—A_ point is an indivisible position. 

3. Def —A line is a position composed of all those through 
which a point, in moving, passes. 

4, Def.—A line is said to be straight, when the point which 
produces it constantly moves in the same direction, which is 
that in which any point of the line lies from one previously 
passed over in the motion. It may be prolonged to any extent, 
since space has no limit. If the point anywhere stops, and 
moves in the opposite direction, that is, towards the points pre- 
viously passed, it will pass them all again, and may also be pro- 
longed in this direction to any extent. It is evident that two 
straight lines which have two points in common entirely coin- 
cide; for if they have two points in common, the direction of 
both lines is that in which one of these points hes from the 


Q. ELEMENTS OF GEOMETRY. 


other: they run from either of these points towards and beyond 
the other in the same unchanging direction, and consequently 
can never separate. 

~ 5. Defi—A line is said to be broken, when it occasionally 
changes its direction. A ‘broken line is composed of straight lines. 

6. Def—A. line is said to be curved, when it constantly 
changes its direction. A curved line may be conceived also as 
composed of straight lines, the number of them being greater 
than any assignable number, or being, what is called in mathe- 
matics, infinite. A curved line is, in this view, a species of 
broken line. 

A curved line is sometimes called simply a curve ; and a 
straight line, simply a line. 

7. Def.— ‘A surface is a position composed of all those 
through which a line, straight, broken, or curved, in moving, 
passes. If, however, the line be straight or uniformly curved, 
it may move along itself, so that the surface will reduce to a 
line. The line which forms a surface may change its form in 
moving, and may increase or diminish at either end or at both. 
It may also break up into two or more lines, or, being broken 
up, it may completely or partly unite. We shall not have, 
however, to consider these cases. 

8. Proprosrrion 1.—J/f a line be moved from any position, 
in any way im which one of tts points remains fixed, till 
it has a position the reverse in direction to its original one, 
there will be. some intermediate position in which the soos 
which it forms with its original and final positions can be 
superposed. 

Let the line AB be moved round A to the reverse position 
AB’, the point A remaining fixed. 

Taking a definite point, C, on the 
o” line AB, C’ being its reversed position, 
there will be some position of the line 
in its movement for which O’C” is 
equal to CC” (C” representing the po- 
sition of C in the intermediate position 
of the line) by the principle of continuity; for at the outset 
CC” is equal to zero, and O’O” equal to C’C, while at the end 
the reverse is the case. 


B” 


B' C’ A Cc B 


ELEMENTS OF GEOMETRY. 3 


_ Now the figure composed of OC and O’’C’ can evidently be 
reversed and superposed on itself, C” falling on itself, C on OC’, 
and C’ on OC; when this is done, the point A, being the middle 
point of O’O, will fall on its previous position; hence C’AC” 
will fall on CAO”, and vice-versa, as was to be proved. 

9. Def—A line in this position is said to be perpendicular 
to the line formed by its first and last directions. In any posi- 
tion in which this is not the case, it is said to be oblique. 

10. Def—If a line AB” be turned ronnd a line B’AB, to 

which it is perpendicular, without changing the perpendicular- 
ity, the surface produced is called a plane. The line which 
turns is called the forming line of the plane; the line round 
which it turns is called the avs of the plane; the point where 
they meet is called its centre. 
11. Def—The amount of turning, or votatzon, as it is called, 
of the forming line in passing from one position to another is 
called the angle between its first and last direction. When the 
forming line becomes perpendicular to its first direction, or has 
made one quarter of a complete rotation, this angle that it 
makes with this first direction is called a rzght angle. The 
angle which it makes with the direction opposite to the first is 
evidently also a right angle. An angle less than a right angle 
is called an acute angle; one greater than a right angle, an od- 
tuse angle. 

12. Prop. 2.—/f one of two straight lines is perpendicular 
to another, the other is also perpendicular to tt. 

Let AC be perpendicular to DCE, then will DC also be per- 
pendicular to ACB. For we can turn 
the figure ACE so that CA will fall on 
CD, and CE on the original position of 
CA, since the figures made by CA with 
CD and CE can (8) be superposed, or, in 
other words, are the same. Then will the 
prolongation of EC fall on CB. But the 
figure formed by CA with this prolon- 
gation of EC in the new position, as well 
as in the original one, is the same as that formed by it with 
CE. Hence the pane formed by OD, on which CA falls, with 
CB, on which the. prolongation of EC falls, is the same as the 





4 ELEMENTS OF GEOMETRY. 


figure formed by it with CA, on which CE falls. Hence OCD 
is perpendicular to ACB. | 

In the same way it can be shown that CE is perpendicular 
to ACB, and that CB is perpendicular to DCE. The lines AB 
and DE are said to be perpendicular to each other. 

13. Prop. 3.— To any line, at any one point of that line, 
only one perpendicular can be drawn in any plane which the 
line can form round that pot as a centre. 

For the part of the line on either side of the point, in form- 
ing the plane, moves from its original direction to the oppo- 
site one, and, after a quarter of a rotation, is (11) in a di- 
rection at equal angles with these two; and as its angle with 
its original direction is constantly increasing, and its angle 
with the opposite one constantly decreasing, there will be 
only one position in which these angles are equal to each 
other. 

The same will be the case as it comes from the opposite di- 
rection back to the original one, and since it has been shown 
(12) that when a line is perpendicular to another, its pro- 
longation in the opposite direction is also perpendicular to 
that other, the single position of perpendicularity on the return 
movement must be the prolongation of the one on the outward 
movement. | 

14. Corollary 1.—Only one plane can be formed round a line 
as an axis at any one point of that line. 

7 Suppose the point to be OC, on the line AB. 

Rotate the line AB round OC, forming a plane. 

Now any two perpendiculars, CD and CE, 

which can be drawn to AB at C, if rotated 
aa round AB, will come into this plane which 
has been formed; and when they do they 
will ‘coincide, by this propasition. Hence, 
in their rotation they will form one and the 
same plane. 

15. Corollary 2.—This one plane includes all the perpen- 
diculars which can be drawn to AB at C; they will all be posi- 
tions of its forming line. 

16. Scholiwm.—The proposition may be more generally stated 
thus: Only one line can be drawn from the point of meeting 


B 


ELEMENTS OF GEOMETRY. 5 


of two lines in any plane of which they are forming lines, so 
as to make equal angles with them. 

17. Def-—A corollary is a consequence easily deduced 
from, and a scholwwm is a remark made on, one or more propo- 
sitions. 

18. Prov. 4.—All right angles are uf 
equal. 

Let CD be perpendicular to AB, and 
GH to EF; then will the angles ACD 4 ¢ - 


and EGH. be equal; by the angles ACD iH 
and EGH we mean the angle from CA 
to CD or from CD to CA, and from GE 


E C F 
to GH, or from GH to GE. 

Place the lines EF and AB in coincidence, the point G fall- 
ing on C. Then rotate CD round these coinciding lines. The 
plane which it forms includes all the possible perpendiculars to 
AB (15); but GH is perpendicular to AB, since EF coincides 
with AB; hence GH lies in this plane, and coincides with some 
position of CD; hence the angles ACD and EGH are equal. 

19. Cor.—Two planes coincide throughout, if their axes and 
centres coincide; or with the same axis and centre, only one 
plane can be formed. 

Schol.—An. angle is usually denoted by three letters, as in 

this proposition ; sometimes, however, when no ambiguity can 
arise, simply by the letter at the point round which it is formed. 

20. Prov. 5.— Two lines which have only one point in com- 
mon, are positions of the forming line of one plane, and of 
only one. 

Let CA and CB be two such lines. A 
Form a plane round © as centre, and 8 
CA as axis, and draw in the plane the 
forming line, CD, at pleasure. Let CE 
be the prolongation of CD in the other E 
direction, or the opposite position of the 
forming line. Now if CB is not per- HF 
pendicular to DCE, the line from one of 
its points, as B, to a point, D, on one of the lines CD, CE, for 
which OD = OB, will be less than that to a similar point, E, on 
the other (9). But if DCE be rotated round CA to the opposite 


6 ELEMENTS OF GEOMETRY. 


position, this will be reversed ; so that, by the principle of con- 
tinuity, there must be some intermediate position, as ICG, in 
which the lines from B to the two points will be equal. CB, 
therefore, will be perpendicular to FCG; but since CA is also 
perpendicular to FCG, FCG will be the axis of a plane of the 
forming line of which CA and CB are positions. 

If CB is perpendicular to DCE, DCE itself will be the axis 
of the plane required. 

Secondly, no other plane can have both these lines CA and CB 
as positions of its forming line. For, if so, the axis of the plane 
must be a forming line of the plane which has CA for its axis, 
since this plane contains all the lines perpendicular to CA at 
C (15). Let this axis, then, be CH for example. Now turn 
the system of lines CF and CB round CA as an axis. CB will 
continually take new positions, by a progressive movement, 
similar to that of CI’, not returning on itself till CF has made 
a complete rotation, and returns on itself. 

When CF falls on CH, and also when it falls on the prolon- 
gation of CH beyond C, CB will (15) be a position of the form- 
ing line of the only plane which can (19) be formed round CH 
at C. But neither of these positions coincides with the original 
position of CB, unless CH. coincides with CF or CG, since, as 
has been remarked, CB only returns on itself when CF returns 
on itself; and there is no other position of the forming line of 
‘this plane round CH which makes the angle with CA which 
the original position of CB makes; hence the original position 
of CB cannot be a position of the forming line of this pa 
round CH, unless CH coincides with CF or CG. 

Or, in other words, CH, if it is different from OF and CG, 
is not the axis of ‘a saree having CA and CB for positions ae 
its forming line. 

There is consequently no line except FCG which is the axis 
of such a plane, and no such plane except one round this axis, 
and the centre C; but round this axis and the centre C, only 
one plane can be formed (19). 

21. Cor.—Any two lines which have a common point will 
form an angle with each other, in the sense in which an angle 
has been defined ; or, in other words, a definite amount of rota- 
tion round the proper axis will bring one of the lines into coin- 


ELEMENTS OF GEOMETRY. f 


cidence with the other. Any two angles, made by two pairs 
of lines, can therefore be added to each other, or one can be 
subtracted from the other. 

22. Def-—The plane thus defined by two meeting lines is 
called the plane of those lines. 

23. Def-—The lines which form an angle are called its sides ; 
the point where they mect is called its vertex. An angle is 
said to be encluded between the two lines or sides which meet 
at its vertex. 

24. Defi—If three points in space be connected by straight 
lines, the figure formed is called a triangle, from its having 
three angles, the vertices of which are at the three points. The 
three lines connecting the points are called the s¢des of the 
triangle ; if all these sides are equal, the triangle is said to be 
equilateral ; if two sides are equal, it is said to be dsosceles ; if 
no two are equal, it is said to be scalene. 

25. Prov. 6.—/f a line be drawn from any point of another 
line, the sum of the two angles which it makes with that other 
line is equal to two right angles. 

For the sum of these two angles is a half-rotation of the 
other line round the axis of the plane of the two lines. But 
this is equal to two right angles. 

26. Cor. 1.—The opposite or vertical 
angles made by two lines which cross each 
other, as the angles ACD and BCE, are 
equal. For ACD + ACE = 2R (Rh rep- 
resenting one right angle); but BCE + 
ACE also = 2R; hence ACD = BCE. | 

This is also clear from the reason that the same rotation 
of the line DCE brings CD on CA, and 
CE on CB. 
9%. Cor. 2.—The sum of the four 
angles formed by the two lines is equal 
to four right angles. And the sum of 4 
ali the angles, ACB, BCD, ete., made 
by any set, CA, CB, CD, ete., of form- 
ing lines of the same plane, is equal 
to four right angles. 

28. Prov. 7.—Two triangles, which have two sides, and the 








8 ELEMENTS OF GEOMETRY. 


included angle of the one equal to two sides and the included 

angle of the other, are equal in all their parts. 
Let the triangles ABC, DEF, have DE = AB, DF = AC, and 
the angles D and A equal; then will 


DEF be equal to ABC in all its parts. 
Place D on A, and DE on AB; then 
A ¢ rotate the plane, having DE and DF for 
E 


forming lines, round the coinciding lines 

DE and AB, till it falls on AC. DF 

will then fall on AC, since the angles 
4 F D and A are equal. 

F will then fall on C, since DF =AC; but E fell on B, since 
DE equals AB; hence EF coincides with BC (4), and is equal 
toit; and the angles E and F are respectively equal to B and C. 

29. Scholium.—In two triangles which are equal in all their 
parts, the equal angles lie opposite the equal sides. Triangles 
which are equal in all their parts are said to be equal, “in all 
their parts” being understood. 

30. Prop. 8.—Jn an isosceles triangle, the angles opposite the 
equal sides are equal. 

Let ABC (Fig. 1) be an isosceles triangle, having AB = CB. 
Reverse it (as in Prop. 1) to the posi- 
tion of Fig. 2. The two triangles, 


B 
\ ABC, CBA, have BC (Fig. 2) = BA 
(Fig. 1); BA (Fig. 2) = BC (Hig. 1), 
and the angles at B equal; hence 
the triangles are equal (28); and the 
y Ae 4 angle A (Fig. 2) = the angle C (Fig. 


1). But the angle A is the same in 
Figs. 1 and 2; hence the angles A and C in Fig. 1 are equal. 
31. Scholium.—Equal triangles will always require a re- 
versal like this that they may coincide, when the equal sides and 
angles come in the re- 
Re PM verse order in the two 
triangles, as one travels 
round both of them from right to left, or from left to right. 
The annexed diagram gives an example of two triangles, 
one of which will have to be reversed before it can be super- 
posed on the other. 


Fig. 1. Fig. 2. 


ELEMENTS OF GEOMETRY. 9 


32. Prop. 9.—A line drawn from the point of meeting 
of the two equal sides of an isosceles triangle to the middle 
point of the opposite side, will be perpendicular to that 
side, and will make equal angles with the equal sides ; and 
wt well be a position of the forming line of the plane of the 
equal sides. | 

Let ADB be an isosceles triangle, having ‘ 
DB=DA. Draw DC to the middle point 
of AB; then will the triangles CAD, CBD, 
have CA = CB, AD = BD, and the angles 
A and B equal; hence (28), the angles ACD 
and BCD are equal (since they lie opposite 4 4 B 
the equal sides DA and DB), and CD is 
therefore perpendicular to AB ; and the angles ADC and BDC 
are equal, since they lie opposite the equal sides AC and BC. 

The point C must lie in the plane of DA and DB. For, if 
not, reverse the triangle ADB, as in Prop. 8, putting DA on 
DB, and DB on DA. The line DC will then make the same 
angle with the farther end of the axis of the plane of DA and 
DB that it now makes with the nearer. Hence, if it fell before 
on one side of that plane, it will now fall on the other. But 
this is impossible, for the point C must have the same position 
that it had before the reversal, since the line AB,of which it . 
is the middle point, occupies its previous position. Hence, the 
line DC cannot fall on one side or the other of the plane, but 
must be one of its forming lines. 

33. Schol.—This demonstration only serves for the point C, 
since no other point of the line AB resumes its previous position. 

34, Prop. 10.—/f lines be drawn from two points to any 
third point, these lines will be equal if this third point rs on 
a line perpendicular to the line connecting the first two ata 
point midway between the first two; and, conversely, of these 
lines are equal, the third point will be on such a perpendicular. 
- Let A and B be any two points (using the figure of Prop. 9), 
and © the point midway between them. 

1. If CD is perpendicular to AB, the triangles ACD and 
BOD are equal (28); hence AD=BD. 2. lf AD=BD, 
the triangle ADB is isosceles; hence the angles A and B 
are equal (30); hence the triangles CAD and CBD are equal 


10 ELEMENTS OF GEOMETRY. 


(28); hence the angles ACD and BCD are equal, or CD is per- 
pendicular to AB. : 

35. Cor. 1.—lf the lmes AD and BD are not equal, CD will 
not be perpendicular to AB; for if it was, by the first part of 
the proposition, AD and BD would be equal; and, conversely, 
if CD is not perpendicular to AB, AD and BD will not be 
equal; for if they were, by the second part of the proposition, 
CD would be perpendicular to AB. 

36. Cor. 2.—By. Prop. 3, Cor. 2, this proposition and its 
corollary can be condensed into the statement that a plane 
described round any line as an axis includes all and only those 
points in space, the lines from any one of which to two points 
on the axis equally removed from the centre are equal. 

37. Prov. 11.—TZo any line, from any point outside, a per- 
pendicular, and only one perpendicular, can be drawn. 

Let BD be the line, and A the point outside. Draw AC to 
any point, C, of the lime BD. Lay off 
in the plane of CA and CD the angle 
DCE = ACD, and make CE=CA.. 
Draw AE. The triangle ACE being 
isosceles, a line from C to the middle 
of AE lies in the plane of CA and 
CE, and makes equal angles with 
them, by Prop. 9; but this is true of 
CD; hence (16) CD coincides with 
this line, and cuts AE at its middle 
point; and hence, by the same Prop. 9, it is perpendicular 
to AE. 

There is no other perpendicular from A to BD. For if any 
other line, as AI’, is perpendicular, prolong it to G, making 
FG = AF, and draw CG. We have, then, CG = CA, or the 
triangle ACG is isosceles, since the triangles CFA and CFG 
are equal (28); hence the line CG is in the plane of CA and 
CD (82). But if so, it must coincide with CE, since the angle 
FCG = FCA, and hence = FCE; and the point G must fall 
on K, since CG, being equal to CA, must also be equal to CE. 
Hence AFG must fall on AE (4); or the supposed new perpen- 
dicular must coincide with the one already drawn. 

88. Prop. 12.—Lines drawn to any line from a point out- 





ELEMENTS OF GEOMETRY. pial 


side make acute angles with it on the side next the perpendicu- 
lar, and obtuse angles on the other side. 

Let AC be perpendicular to CB; then will “ 
ABC be an acute angle. It cannot be a right 
angle, by Prop. 11; and if it could be obtuse, 
some point, D, could be found on AC, by the 
principle of continuity of motion, for which 
DBC would bearight angle. But this would " 2 
be equally contrary to Prop. 11. The angles on the other side 
are obtuse (25). 

39. Cor. 1.—The angles opposite the equal sides of an isosceles 
triangle are acute, for the perpendicular falls half way be- 
tween them. 

40. Cor. 2.—A triangle cannot have two obtuse angles, or 
one right and one obtuse angle; in short, it must have at least 
two acute angles. | 

41. Prop. 13.—1. The perpendicular is the shortest line 
which can be drawn from a point to a line; 

2. Two lines equally removed from the perpendicular are 
equal ; 

38. Of two lines unequally removed from the perpendicular, 
the nearer is the shorter. 

Let AC be perpendicular to BF; it A 
is shorter than any other line, AD, 
from A to BF. It cannot be equal to 
AD, otherwise the triangle CAD would 
(30) have the angle ADC a right angle ; 
but this cannot be (37); or the same ¥F c Aware 
may be deduced from Prop. 12, Cor. 1. 
Neither can it be greater than AD. 
For, if so, lay off AE = AD. By Prop. 12, Cor. 1, the angle 
AED is then acute; but, by Prop. 12 itself, it must be obtuse. 

2. The equality of the lines AB and AD, for which CB and 
CD are equal, results immediately from Prop. 7, as in (84). 

8. AD < AF. For, produce AD, and drop a perpendicular 
to it from F, ADF is obtuse (38); hence this perpendicular 
will fall at a point, G, beyond D (by the same proposition). Now 
AG < AF, by the first part of the present proposition ; hence, 
still more, AD < AF. 


D 


12 ELEMENTS OF GEOMETRY. 


42. Prop. 14.—Any side of a triangle ¢s shorter than the 
sum, and longer than the difference of the other two. 

Let ABC be a triangle, in which the angle B is acute, O 
acute or obtuse, as represented in the 
figure, the perpendicular AD from A 
falling in the first case on the side BC, in 
the second on the same side produced. 
B C pc We have (41) BD < BA, and CD < CA; 

hence BC or (BD + CD) < BA + OA, the 
upper sign referring to the first case, the lower to the second. 
The same can be shown of any other side, since (40) at least 
one of the angles at its extremities must be acute. 

From the inequality BC < CA + BA, we have BA > BC 
— CA; and from the inequality CA < BC + BA, we have 
BA > OCA — BC. 

43. Schol.—A triangle can always be constructed with three 
given limes as sides, if one of these lines is less than the sum 
and greater than the difference of the other two. For place 
these other two end to end, so as to form one straight line; 
the lines connecting the ends which do not touch will be equal 
to the sum of the two lines. Now turn one of the lines round 
the ends which touch each other, till its direction is reversed, 
so that it coincides with the other, any course whatever being 
followed during the movement. At the end of the movement 
the line connecting the other ends of the lines will be equal to 
the difference of the two. Hence, by the principle of contin- 
uity, some intermediate position of the moving line can be 
found for which the line connecting the other ends will have 
any required length, intermediate between the sum and the 
difference of the two; it being equal to their sum at the begin- 
ning of the movement, and to their difference at the end. In 
some position, therefore, it will have a length equal to that of 
the third line, and will in this position form with the other two 
the triangle required. 

If one of the sides of a proposed triangle does not con- 
form to the above conditions, it is evident, by the proposition, 
that the triangle cannot be constructed. 

44. Prop. 15.—A straight line is the shortest line which can 
be drawn between two points. 


A 


ELEMENTS OF GEOMETRY. 13 


1. The straight line AB is shorter than any broken line 
between A and B,as ACDEB. For (42) 
AD < AC + CD; AE < AD + DE, 
AB < AE + EB; hence, still more, 
AB < AC + CD + DE + EB. 

2. It is shorter than any curve be- 
tween A and B; for a curve is merely 
a line broken to an indefinite, or to what is called in mathe- 
matics, as has heen remarked (6), an infinite extent. 

45. Def—The straight line between two points calledsthe 
distance of these points from each other; the perpendicular 
from a point to a line is called the fig ew ot the euint from 
the line, or of the line from the point. - 

46. oa 16.— Two triangles, having two 
one equal to two sides of the other, and the incl™ 
equal, will have the third sides unequal; and th® 










Let ABC be the triangle which has the greater includé 
angle, A; a point, D, can (by the 
principle of continuity) be found on 8 


the line BC, such that the line AD Er 

shall make with AC an angle equal CG 
to the included angle in the other LF 
triangle. There are three cases, ac- 4 ures 


cording as the line AD is less than, 
equal to, or greater than AB, or the equal side in the other 
triangle. The cases are denoted by (1) (2) (8) in the figure. 
Lay off on AD (prolonged of course in the first case, AE = AB. 
In the second case D and E coincide. The triangle AEC 
being equal (28) to the triangle which has the smaller included 
angle, we have only to show that AE + EC < AB + BC. 

For the first case, we have (42) AB < AD + BD, and 
EC < DE + DC. Adding these inequalities, and putting in 
the first member for AB its equal AE, and in the second mem- 
ber for AD + DE (= AE) its equal AB, we have AE + EC < 
AB + BC. 

For the second case, we have, manifestly, EC < BC. Add- 
ing AE = AB to this inequality, we have AE + EC < AB 
+. BOC, 


14 ELEMENTS OF GEOMETRY. 


For the third case, we have (42) AD < AB+ BD. 
Adding DC to both members, we have AD + DC < AB + 
BO. But (42) DC > EC — ED; hence, still more shall we 
have (AD — ED) + EC = AE + EC < AB + BC. 

47. Cor.—lf two triangles have two sides of the one equal 
to two sides of the other, and the third sides unequal, the 
angles included between the equal sides will be unequal ; and 
the smaller angle will belong to the triangle which has the 
shorter third side. 

For if this angle is not smaller, it must be equal to, or 
greater than, the corresponding angle in the other triangle; 
but it cannot be equal to that angle, for if so (28), the third 
sides would be equal; nor can it be greater than that angle, for 
if so, the third side opposite would be greater than the third 
side in the other triangle, by the present proposition. 

48. Prop. 17.— Two triangles having the three sides of one 
equal to the three sides of the other, are equal in all their parts. 

For if the angle between any two sides in one triangle is less 
or greater than the angle between equal sides in the other tri- 
angle, the third side will (46) be less or greater than the third 
side in the other triangle, but this is not the case; hence, the 
angle between any two sides in one triangle must be equal to 
the angle between equal sides in the other triangle. 

49, Prop. 18.—A line connecting any two points of a plane, 
lies wholly in that plane. 

Let AB be two points in the plane, the centre of which is 
C, and axis DCE. Lay off CE = OD. 

We have then to show that any point of 
the line AB, as F, for instance, will lie in the 
plane. Draw CA, DA, EA, OB, DB, EB, 
CF, DF, EF. By (84) or by (41) we have, 
‘ ” since C is the middle point of DE, AD) Se, 

and BD = BE. 
Hence, the triangles ADB, AEB, are 
equal (48); hence, the angles DAB and 
EAB are equal. 

Hence, the triangles DAF, EAF, are equal (28); nent 
Dee EF, 

Hence, since C is the reales point of DE, the line CF will 


DB 


ELEMENTS OF GEOMETRY. 15 


be perpendicular to DE (84); or, in other words, the line 
CF is a forming line of the plane. Hence, F lies in the plane. 
The same reasoning will apply if F is in AB prolonged. 

50. Cor.—The angle ACB = ACF + FOB, since CA, CF, 
‘and CB are in the same plane round C; and generally, if any 
number of lines are drawn to a line from a point outside, the 
angle between any two of them will be equal to the sum of the 
partial angles formed by the successive intermediate lines. 

51. Prov. 19.—TZwo triangles having two angles and the 
included side equal, are equal in all their parts. 

Let ABC be one of the triangles, having A, C, and AC the 


the same as two angles and the included 
B 


side of another triangle. D 
Lay off on AB, produced if necessary, 
a distance, AD, equal to the side opposite , 


the angle equal to C in the other triangle. 

Join DC. The triangle ADC will be equal (28) to the other 
triangle; hence, DCA = BCA. But this is impossible (50) 
unless the point D fallson Bb. But if it does, the other tri- 
angle is equal to ABC in all #ts parts, which was to be proved. 

52. Prov. 20.—A triangle having two angles equal, is isosce- 
les, having the sides opposite its equal angles equal. 

This is proved from Prop. 19, just as Prop. 8 is proved from 
Prop? 

53. Prop. 21.—Jln any triangle having two unequal angles, 
the smaller of these two angles has a shorter side opposite it 
than the other has; and conversely im any triangle having two 
unequal sides, the shorter of these sides has a smaller angle 
opposite it than the other has. 

In the triangle ABO, let the angle C be smaller than the 
angle A. Lay off the angle DAC equal to the angle C; DAC 
will, by Prop. 20, be an isosceles trian- 
gle, having AD = CD. ButAB< AD & 
1B (42); hence AB < CB. 
Conversely, if in the triangle ABC 
‘we have AB < CB, we shall have the 
angle © less than the angle A. 

For if it were equal, we should have (52) AB = CB; and 
if it were greater, we should have, by the present proposition, 


A Cc 


16 ELEMENTS OF GEOMETRY. 


AB > CB; hence it must be less, since AB is neither equal 
to, nor less than CB. j 
54. Cor. 1.—The exterior angle, BAE, of a triangle, ABO, 
made by one of the sides with the prolongation of another, is 
always greater than the angle of the 
B triangle BCE, which this prolonged 
side makes with the third side. 
When the perpendicular on the pro- 
longed side from the opposite angle 
falls between the two other sides, as in 
EA, A, ¢ BA,,C, the exterior angle is obtuse, 

while that of the triangle is acute; so 
that in this case the truth of our proposition is evident. When 
the two sides are on the same side of the perpendicular, as in 
BA,C, draw BA, as far on the other side as BA, is on the side 
on which BC lies. Then, by Prop. 18 and the present propo- 
sition, the angle BA,C > BCA,; but BA,C = BA,E; hence, 
BA,E > BCE, BCE being the same as BOA, Also BOF = 
2R — BCE > BA,F = 2R — BAE. 

55. Cor. 2.—The angle made with any line by a line drawn 
to it from a point outside, approaches a right angle as the line 
approaches the perpendicular. 

56. Cor. 3.—If a triangle has an angle and one of the in- 
cluding sides equal to an angle and one of the including sides 
of another triangle, and the other including side greater or less 
than the other including side in the other triangle, the angle 
opposite the equal including side will be less or greater respect- 
ively than the angle opposite the equal including side in the 
other triangle ; and the converse will be true; that is, we may 
interchange in the above statement “ other including side”’ and 
“angle opposite the equal including side.” 

57. Defi—A triangle is said to be rzght-angled, or simply 
right, when one of its angles is a right angle. 

58. Def—The side opposite the right angle is called the 
hypothenuse. 

59. Schol.—There is only one right angle (87), and conse- 
quently only one hypothenuse. The other angles are acute (88). 

60. Prov. 22.—T'wo right triangles, having the hypothenuse 
and another side equal, are equal in all ther parts. 


ELEMENTS OF GEOMETRY. 17 


Let ABC be a triangle, right-angled at B; and suppose 
another triangle, also right-angled, and having an equal hy- 
pothenuse and a side equal to AB. Place 
this equal side upon AB, and revolve the 
other side round the coinciding sides till 
it takes the direction BC. If it is longer 
or shorter than BC, the hypothenuse in 2 A 
this triangle will (41) be longer or shorter 
than AC. But this is contrary to our supposition ; hence, this 
side = BO, and the triangles are equal (28). 

61. Prop. 28—Two right triangles, having the hypothenuse 
and one of the acute angles equal, are equal in all their 
parts. | 

Let ABC be a triangle, right-angled at B; and suppose 
another triangle, also right-angled, and having an equal hy- 
pothenuse and an angle equal to C. j 

Lay off-on CB, produced if necessary, a 
distance, CD, equal to the side adjacent to 
the angle equal to C in this triangle; and 
‘draw AD. ADC will (28) be equal to 2; c 
the triangle. But this triangle is right- 
angled; hence ADC will be a right angle. But this is impos- 
sible (87) unless D falls on B; and if it does the triangles are 
equal in all their parts, which was to be proved. 

62. Prop. 24.—A right triangle, having an equal hypothe- 
nuse with another, and a smaller acute angle, has a smaller 
- side opposite that angle, and a greater side adjacent to it, than 
the other triangle has ; and its other angle ts greater than that 
of the other triangle. 

Place the hypothenuse of one on that 
of the other, and let the smaller angle be B 
BAO, the larger, BAD. Let the triangle 
BAC be turned round AB till AC cuts 
BD in E, between B and D (50). A c 

Reo < BD: and :-BC < BE (41). D 
Hence, still more, “BC << BD; 

2. The point E will be between A and C; otherwise we 
should have AED obtuse. Hence, AC > AE; eit AE > AD 
(41). Hence, still more, AC > AD. 

2 


A 


18 ELEMENTS OF GEOMETRY. 


3. And,since E is between A and C, the angle ABC > ABD 
(50). 

63. Cor—aA right triangle having an equal hypothenuse 
with another, and a smaller side, BC, has a smaller angle 
opposite that side than the other triangle has. For if the angle 
were equal, the side would (61) be equal; and if the angle 
were greater, the side would be greater, by this proposition. 
Hence, the angle is smaller, as was to be proved. Hence, we 
shall also have in this case, the other side and angle greater 
than the other side and angle respectively, in the other triangle, 
by this proposition. 

64, Prop. 25.—The three planes formed at the vertices 
of the angles of a triangle as centres, by the three sides, are 
edentical. | 

Let ABC be the triangle. 

Prolong any two of the sides, as AB 
and CB, beyond B, to any convenient 
distances, as BD and BE, and draw 
CD, DE, and EA. 

The plane round A includes CB (49), 
and hence, the point E lies init. And 
hence, also, by the same, the line DE lies in it; and the same 
is true of CD, and also, of course, of AC and AE. And hence, 
by this same proposition, every line drawn from B to any point 
of the figure ACDEA lies in the plane round A. 

But since the plane round B includes this whole figure, by 
the same proposition, these lines drawn from B make up the 
whole plane round B; hence (since these lines all connect B 
with another point in the plane round A), every point in the 
plane round B lies in that round A. 

It can be shown in the same way that every point in the 
plane round A lies in that round B; hence, these two planes 
are identical; and the same can be shown for the planes round 
A and OQ, or round B and C. 

65. Cor—Through three points, not lying in the same 
straight line, only one plane can be passed. For any plane 
passed through the three points includes the lines connecting 
them (49), and hence, it is, identical with the three planes 
treated of in the present proposition. 





ELEMENTS OF GEOMETRY. 19 


66. Prop. 26.—The plane formed by any two imtersecting 
lines of another plane, is identical with that other plane. 
Let AA, BB, be two intersecting 


lines lying in a plane round K; and : B 
let a plane be formed passing through 4 je 
them and centred at their intersection, * 

C. This plane will be identical with D~, 


that round K. 

Take a point, D, at pleasure, on that part of one of the lines 
which lies on the opposite side of the other from that on which 
K lies. Draw KO, KD. KD is a forming line of the plane 
round K, and therefore intersects CB at some point, E. The 
plane round C is the same (64) as that of EC and ED, or of 
EC and EK; but this is the same as that of KO and KE, or 
that around K, by the same. Hence, the planes around C and 
K are identical, which was to be proved. 

67. Cor.—Hence, any point of a plane will serve as a centre ; 
and for every new centre there is of course a new axis. 


3 © © Koh 


THEORY OF PARALLELS. 


68. Def:—Lines which are axes of the same plane are 
said to be parallel. It is plain that through any point a par- 
allel can be drawn to a given line, since a plane can be formed 
by rotating a perpendicular from the point to the line, round 
the line. 

69. Prop. 27.—Two parallel lines can never meet. 

For if they did they would be two perpendiculars from their 
point of meeting to the line joining them in the plane; but 
this is impossible (37). 

70. Prop. 28.—A line which ts parallel to another, lies in the 
plane formed by the other and any line connecting the two. 

Let AB be parallel to CD; then 
will AB lie in the plane of DO and 
DB; Bb and D being any two points 
on the lines AB and CD. Let AC be 
the line connecting AB and CD in the 
plane of which they are axes. Drawa 
line through C in the same plane, and 
perpendicular to CA, and lay off on it 
equal distances CE and CF. Draw AE, AF, BE, BC, and BF. 

The lines AE and AF are equal (41); hence, since BAE and 
BAF are right angles, the triangles BAE and BAF are equal 
(28), and hence, BE = BF. Hence (86), CB lies in the plane 
round EF at C, that is to say, in the plane of CA and CD, or 
CA in that of CD and CB, which (64) is the same as that of 
DC and DB. Hence, A is in the plane of DC and DB; and 
of course B is in that plane ; hence (49), AB is in that plane. 

71. Cor. 1.—I£ we have two lines perpendicular to each 
other, and a third line in the plane of the two, and perpen- 





ELEMENTS OF GEOMETRY. : 21 


dicular to one of them, this third line will be parallel to the 
other. 

Let CD and CA be the first two lines, and AB the third. 
Erect at A an axis of the plane formed by CA round CD as an 
axis; this axis at A will, by the present proposition, be in the 
plane of CD and CA, and it will, since it is an axis of the plane 
formed by CA, be perpendicular to CA. But only one perpen- 
dicular to CA at A, in the same plane, is possible (18); hence 
AB, which is perpendicular to CA, must be this axis at A, and 
hence it is parallel to CD. 

72. Cor. 2.—From any point, B, outside of a plane, an axis 
(or perpendicular, as it may also be called) can be drawn to 
that plane; and only one can be so drawn. For, draw an axis, 
CD, to the plane at any point, C, of the plane; if it does not 
pass through the point B outside, draw a line, CA, in the plane 
which CD forms with the point B, perpendicular to this axis at 
its centre, C; this line will lie in the first plane on account of 
its perpendicularity to the axisCD. Nowdrop on this new linea 
perpendicular, BA, from the point outside. This perpendicular 
will, by Cor. 1, be parallel to OD, and hence will be an axis or 
perpendicular to the first plane. Only one can be drawn, by 
Prop. 27, 

73. Prov. 29.—If a perpendicular be dropped from any 
point on a line lying in any plane, and an axis be drawn to 
the plane through the point from which the perpendicular is 


_ dropped ;. any line connecting a point of the axis with the 


foot of the perpendicular will also be perpendicular to the line 
of the plane. 

For, let B be the point, and EF the line of the plane, using 
the figure of the last proposition ; and let BC be perpendicular 
to EF; BE and BF will be equal (41), as in the last propo- 
sition; and if AB is the axis through B, AK and AF must 
also be equal (60); and hence (28), GE and GF, drawn from 
any other point, G, of the axis AB will be equal, and GC will 
(34) be perpendicular to EF. (GC, GE, and GF are not drawn, 
for the sake of simplicity in the figure.) 

74. Cor—lf two points, G and G’,be taken on the axis of 
any plane, at equal distances from the centre, they will be 
(41) equally distant (45) from any line drawn in that plane. 


22 ELEMENTS OF GEOMETRY. 


75. Prop. 380.—The sum of the angles of a triangle vs not 
greater than two right angles. 

Let ABA’ be 
the triangle. 
Draw at A’a line 
A’B', equal to 
AB, in the plane 
of A’A and A’B, 
and making an 
angle B’A’C with the prolongation of AA’, equal to the angle 
BAA’. This line will fall between A’B and A’C, since (54) 
the angle BA’C is greater than BAC. 

We have AA’B + BA’B’ + B’A’C = 2R; but the sum of 
the angles of the triangle is equal to AA’B + A’BA + B’A’C; 
hence, if this sum be greater than 2R, A’BA must be greater 
than BA’B’; and hence, drawing BB’, we have (46) AA’ > 
BB’; and hence an integer, which we may denote by n, may 
be found so large that n(AA’ — BB’) will be greater than any 
line whatever, for instance 2AB; if, then, the construction 
ABBb’A’ be made 7 times on the straight line AC, and in the 
plane of BAC, we shall have AA” > 2AB + nBB’; or the 
straight line AA” greater than the broken line ABB’B”..... 
BOA, 

But this (44) is impossible; hence it is impossible that the 
sum of the angles of the triangle ABA’ should be greater than 
two right angles. 

76. Cor. 1.—Hence, the difference between the angles BAKE 
and BCE, made by the two lines AB and CB with CA, in 
Prop. 21, Cor. 1, is at least as great as the angle ABC between 
the two lines. 

For ABC + BAC + BCE is not greater than 2h, by what 
has just been shown; hence, ABC is not greater than (QR — 
_ BAC) — BCE, which is the same as BAE — BCE. 

77. Cor. 2.—The sum of the angles of a figure of four sides 
lying in the same plane cannot be greater than four right 
angles; for, if a line be drawn connecting either pair of the 
opposite vertices of the figure, two triangles will be formed, 
the sum of the angles of which will be equal to the sum of the 
angles of the figure. 


B B! Be b(n) 


A A' A" Ary ¢ 


ELEMENTS OF GEOMETRY. 23 


78. Cor. 3.—Hence, if in such a figure, three of the angles 
are right angles, the fourth cannot be obtuse. 

79. Prop. 31.—Jf two lines be drawn in space, not lying in 
the same plane ; two lines connecting them cannot be drawn so 
as to be perpendicular to both. 

Let AB and CD be two lines not in the same plane, and let 
AC and DE be two lines connecting them; we have to show 
that if BAC, ACD, and AED are right angles, CDE cannot 
be a right angle. 

Draw through D (72) an axis to the plane of AB and AQ, 
meeting the plane in F. Draw FC, 

FE, and EC; FC will (73) be perpen- 

dicular to AC, and FE to AB; hence, D 

the angles FCE and FEC will be acute ; 
and hence, if we draw FG perpendicu- 
lar to CE, it will fall between C and E 

E. Draw DG;; it also (73) will be per- A C 
pendicular to CE. 

We shall then have in the triangles DGC and FGC, GC 
common, and DG greater (41) than IG ; hence (55), the angle 
CDG is less than CFG. Similarly, EDG is less than EFG. 
Hence, CDE is less than CFE. But CFE is (78) right or 
acute ; hence, CDE must be acute, and not a right angle, which 
was to be proved. 

80. Schol—I£ DE falls on BA, produced beyond A, 
F will, since EF and AC are (71) parallel, also fall on FC 
produced beyond OC (69); and the same demonstration will 
apply. 

81. We shall now make an assumption, the truth of which 
may be considered self-evident, as it seems to be connected 
with, or to form a part of the very idea of direction, which was 
mentioned in the beginning as one of the three fundamental 
ideas of geometry. 

This assumption is that two lines which do not coincide, may 
still have the same direction; or, more fully, that from any 
point of space a line may be drawn in the sume direction with 
any given line. 

82. Prop. 32.—Jf two lines have the same direction : 

1. They must lie in the same plane ; 


& 


94 ELEMENTS OF GEOMETRY. 


2. Any line connecting them and perpendicular to one, 
must also be perpendicular to the other. 

1. Let AB and CD be two lines in 
the same direction, and suppose them 
not to lie in the same plane; we may 
(79) draw a line connecting them 
which will be perpendicular to one 
and oblique to the other. Let this line 
be EF, perpendicular to AB, but ob- 
lique to CD, making CEF acute, and DEF obtuse. Round 
EF as an axis, and F as a centre, describe a plane; AB will 
lie in this plane. The other axes dropped from the line CD 
on this plane will (70) form a plane cutting the former one in 
a line GH. Now draw a line C’ED’ in this last plane, making 
the angle C’EF = CEF and D’EF = DEF. The line C’ED’ 
or O’D’ has precisely the same position relatively to BA that CD 
has to AB. Hence, if CD has the same direction as AB, C’D’ 
_ has the same as BA, or D’C’ has the same direction as AB, and 
hence the same as CD. But this is manifestly not the case, 
since it intersects CD. Hence CD cannot have the same di- 
rection as AB. | 

2. CD and GH, which lie in the same plane, cannot have the 
same direction if EF’, which is perpendicular to GH, is oblique 
to CD, since in this case also it would follow that D’C’ would 
have the same direction as CD. 

83. Prop. 38.—Parallel lines have the same direction ; and 
conversely, lines which have the same direction are parallel. 

Let AB and CD be two parallel lines. They 
are both perpendicular to AC, connecting them 
in the plane of which they are axes, and they lie 
(70) in the plane formed by either of them with 
AC. Now if AB has not the same direction with 
CD, some other line AE drawn in this plane 
A ¢ must have that direction, in accordance with the 

assumption made, and by the first part of Prop. 
32; but this is impossible, by the second part of that propo- 
sition, since CA must (13) be oblique to AE. 

The converse is evident, by.the two parts of that proposition 
together with Prop. 28, Cor. 1. 





B > D 


ELEMENTS OF GEOMETRY. 25 


84. Cor. 1 —Through any point, only one parallel can 
be drawn to any given line, by the first part of this propo- 
sition. 

85. Cor, 2.—A line which is parallel to one of two parallels, 
is parallel also to the other (by the same). 

86. Def—We may now, if we please, define parallel lines as 
lines having the same eae 

87. Prop. 34.—TLhe sum of the angles of a triangle is equal 
to two right angles. 

First, let us suppose the triangle to be right-angled. 

Let ABC be a triangle right-angled at B. Draw CD parallel 
to BA, and drop AE perpendicular to CD. The angle BAH 
will be (82) a right angle; hence the sum 


of the angles in the two triangles ABC D 
and AEC is equal to four right angles. 
But the sum of the angles in ABC cannot 4 E 


be greater (75) than two right angles; 
neither can the sum of the angles in AEC 

be greater than two right angles; hence, 2 
each sum is equal to two right angles. 

Secondly, suppose the triangle not to be right-angled. 

Let ABC be such a triangle. It must (40) have two acute 
angles; let A and C, then, be acute. Drop BD perpendicular 
to AC, and falling between A and OC, 
since they are both acute. The sum of B 
the angles in each of the right triangles 
thus formed, is equal to two right angles, 
as has been shown; hence the sum of A D C 
the angles in ABC is equal to four right 
angles, less the sum of BDA and BDC; but this last sum is 
equal to two right angles; hence the sum of the angles in ABC 
is equal to two right angles. 

88. Cor. 1.—The Handi angle BAKE, of Prop. 24, Cor. 1, 
there shown to be greater than the angle BCA of the triangle, 
is equal to the sum of the two angles BCA and ABC, which 
are not adjacent to it. 

89. Cor. 2.—The sum of the two acute angles of a right tri- 
angle is equal to one right angle. 

90. Cor. 3.—If two angles of one triangle are equal to two 


cS 


26 ELEMENTS OF GEOMETRY. 


angles of another triangle, the third angles are equal, and the 
triangles are mutually equiangular. 

91. Def—A polygon is a figure formed by a broken line, 
starting from any point and returning to the same point again. 

In the polygons we shall have to consider, the broken line © 
lies in one plane and nowhere crosses itself. 

The number of angles and of sides in any polygon is evi- 
dently the same. 

92. Def—The lines connecting angles of a polygon with 
are not adjacent are called Vien Ra 

93. Def—-A polygon of four sides is called a guadrilateral ; 
of five, a pentagon, of six, a hewagon ; of seven, a heptagon ; 
of eight, an octagon ; of ten, a decagon ; of twelve, a dodeca- 
gon. The others might be similarly named, but are not often 
enough in use to require it. 

94, Prop. 35.—The sum of the angles of any polygon is 
equal to twice as many right angles as the polygon has sides, 
less four right angles. 

For if we draw lines to the angles of the polygon from some 
point within, as I’, we shall form as many 


‘ C triangles as the polygon has sides; and 
A the sum of the angles of the triangles, 
which is therefore twice as many right 
p angles (87) as the polygon has sides, will 
E 


exceed the sum of the angles of the 
polygon by the sum of the angles round F’, which belong to the 
triangles, but not to the polygon. But the sum of the angles 
round F is equal to four right angles. Hence the sum of the 

angles of the polygon is as stated in the proposition. 
95. Scholium.—This proof can be extended to the case of a 
polygon like that inthe annexed figure, in which it is impos- 
sible to find a point within from which lines 
can be drawn to all the angles without cutting 
the sides; but it is not necessary for our 

present purpose so to extend it. 

96. Cor. 1—The sum of the exterior angles 
formed by prolonging the sides AB, BC, CD, 
DE and EA, or AE, ED, DC, CB and BA, is 
equal to four right angles. or, taken together with the inte- 


ELEMENTS OF GEOMETRY. QT 


rior angles, they make two right angles taken as many times 
as the polygon has sides, or twice as many right angles as the 
polygon has sides. 

97. Cor. 2.—The sum of the angles of a quadrilateral = 4R; 
of a pentagon, 6R; of ahexagon, SR, etc. The general for- 
mula for this sum may be written, as given in the proposition, 
2nR — 4K, in which nv is the number of sides, or 2 (n — 2) R, 
or mR + (n — 4) R. 

98. Def-—If a line cut two others lying in the same plane, 
the angles lying on opposite sides of the cutting line, and inside 
the other two, are called alternate wmterior angles; those on 
opposite sides of the cutting line, and outside the other two, 
are called alternate exterior angles. The angles lying on the 
same side of the cutting line, and one inside, the other outside, 
of the other two, are called opposite mtervor and exterior 
angles. 

99. Prop. 36.—/f a line cut two parallel lines, any two al- 
ternate angles, whether interior or exterior, and any two 
opposite mterior and exterior angles, are equal; and con- 
versely, if a line cut two others lying in the same plane, making 
any two such angles equal, the lines cut are parallel. 

1. Let AB and CD be two parallels, and EF a line cutting 
themin Gand H. Drop GK perpen- 
dicular to CD from G; the angle 
KGB will be (82) a right angle, and 
hence we shall have KGH + HGB 
equal to one right angle. But KGH 
+ GHK also is equal (89) to one 
right angle; hence, HGB = GHK. 
Hence also (25), HGA = GHD. 

From the first of these equations we have (26) EGA = 

DHF or GHK; from the second we have by the same, EGB 
= KHF or GHD. 
_ 2. Conversely, if any of these pairs of angles are equal, we 
shall have (25)(26) HGB=GHK. But KGH + GHK is 
equal (89) to one right angle, if GK is perpendicular to CD; 
hence KGH + HGB, or KGB, is a right angle and the lines 
AB and CD are (71) parallel. 

100. Cor—Of the eight angles formed by a line cutting two 





28 ELEMENTS OF GEOMETRY. 


parallels, four are equal acute angles, the rest equal obtuse 
angles, and each obtuse angle is equal to 2h minus one of the 
acute angles, unless one of the angles is a right angle, in which 
case all eight will be so. Conversely, if a line cut two others 
in the same plane, making two angles with them which are 
not alternate or opposite interior and exterior, have a sum equal 
to two right angles, the lines cut are parallel. 

101. Prop. 87.--1. The perpendiculars connecting two par- 
allels are everywhere equal. 

2. Lf two lines in the same plane be connected by two equal 
perpendiculars, either dropped on the same line or one on 
one, and the other on the other, the lines are parallel. 

1. Let EF’, GH be lines connecting two parallels and per- 

pendicular to both of them ; then will 

E ¢ eB they be equal. Draw EH. In the tri- 

E angles EF H, EGH, we have the angles 

: EHF and GEH equal (99) ; hence (61), 

. oF HL © the triangles are equal, and EF = GH. 

2. Let EF and GH. be equal, and 

first suppose the angles EF H and GHF to be right angles. 

Draw EH as before. We have EHF + EHG = R; but EHF 

+ FEH = R (89); hence, EHG = FEH. Hence, the triangles 

EHG and FELT are equal (28); hence, G is a right angle, and 
the lines AB and CD are parallel. 

Secondly, suppose IF’ and G to be right angles. Draw EH as 
before. The triangles GEH and EHF are equal (60); hence, 
the angles GEH and EHF are equal, and the lines AB and CD 
are (99) parallel. 

102. Cor. 1.—The distances EG, FH between the perpendic- 
ulars connecting two parallels are equal, for these perpendic- 
ulars are themselves parallel. Hence also, if equal distances 
EG, FH. be laid off in the same direction on two parallels from 
the extremities of a perpendicular EF connecting them, the 
line GH connecting the extremities of these distances will be 
perpendicular to the parallels. For if not, we could draw a 
perpendicular GL; and we should have FH and FL both equal 
to EG, which is impossible. 

103. Cor, 2.—If two planes be drawn to the same axis at 
different points, all the axes let fall from one on the other will 





ELEMENTS OF GEOMETRY. 29 


be equal to the portion of the common axis to which they are 
both drawn, which is included between them, by the first part 
of Cor. 1, and hence equal to each other; and hence, by the 
second part of Cor. 1, all the lines connecting these axes in the 
plane from which they are dropped will be perpendicular to 
them; hence, these axes are also axes of this plane. 

104. Def—Planes having a common set of axes are called 
parallel planes. It is plain (72) and (14) that through any 
point outside of any given plane, one, and only one plane can 
be passed parallel to the given plane. 

105. Def—A. Parallelogram is a quadrilateral, two opposite 
sides of which are equal and parallel. 

106. Prop. 38.—l/n a parallelogram, the other two opposite 
sides are also equal and parallel, and the opposite angles are 
equal. 

Let AB and CD be equal and parallel. Draw AD. The 
triangles BAD, CDA are equal, (99) and (28); hence BD = 
AQ, and the angles B and © are equal. The angles BAC and 
CDB are also equal, being the sums of equal angles. 


A B A B 
107. Cor. 1.—The diagonals of a parallelogram divide each 
_ other into equal parts, or bisect each other. For the triangles 
AEB, DEC are equal (99) and (51); hence, AE = DE, and 
ie OR. 

108. Cor. 2.——-The sum of any two adjacent angles of a 
parallelogram is equal to two right angles. 

109. Cor. 3—If one angle of a parallelogram is a right 
angle, all the rest are right angles. Tor, by this proposition, 
the opposite angle is a right angle; and the sum of the four 
angles is equal (97) to four right angles; hence, the sum of the 
two remaining angles is equal to two right angles; but they 
also, being opposite, are equal; hence, each is a right angle. 

110. Def—A parallelogram of this kind is called a rectangle. 


If all the sides are equal, that is, if the length taken on the 
parallel lines is equal to the distance between them, it is called 


80 ELEMENTS OF GEOMETRY. 


a sguare. If the sides of a parallelogram are all equal, but the 
angles not right angles, it is called a rhombus. 

111. Prov. 89.—// a quadrilateral has both pairs of oppo- 
site sides either equal or parallel, or both pairs of opposite 
angles equal, it 1s a parallelogram. 

Suppose in the figure of Prop. 88, AB = OCD, and AC = BD. 
The triangles ABD and DCA are (48) equal ; hence, the angles 
BAD and ADC are equal, and hence (99), ABDC is a parallel- 
ogram. 

Secondly, suppose AB and CD, and also AC and BD, to be 
parallel. The angles BAD and CDA are equal, and also BDA 
and CAD (99); hence (51), the triangles BAD and CDA are 
equal, and AB = CD; hence ABDC is a parallelogram. 

Thirdly, suppose the angles B and C are equal; also CDB 
and BAC. The sum of the angles of ABDC = 4R (97); 
hence, the sum of B (or C) and CDB, also of C (or B) and 
BAC, = 2R. Hence (100), AB and CD are parallel, and so 
are AC and BD; hence, by the second part of this proposition, 
ABDC is a parallelogram. 


resto Md ery 3 EB 


SIMILARITY OF FIGURES. 


112. Prov. 40.—A line connecting the middle points of two 
sides of a triangle is parallel to, and half as long as, the third 
side. 

Let D be.the middle point of the side AB of ABC. Draw 
DE parallel to BC. It will lie somewhere , 
between DA and DO, since (69) it cannot A 
meet BOC, nor coincide with DA, which, 
when produced, meets BC. That is, it will 
cut AC in some point E between A and C. 2 E 

Draw similarly DF parallel to AO, and 
cutting BO in F. The triangles ADE, 

DBF are (51) equal, since AD= DB by 8 F c 
hypothesis, and the angles including these 

sides are (99) equal in the two triangles. Hence AE = DF, 
and DE = BF. 

But DECF is (111) a parallelogram, since both pairs of oppo- 
‘site sides are parallel. Hence (106), EC = DF and DE= FC. 

Hence, in the first place, AE = EC, since each is equal to . 
DF; or, DE, which has been drawn from D parallel to BC, is 
the line between the middle points of AB and AC; or, revers- 
' ing the statement, this line between the middle points of AB 
and AC is parallel to BC. 

In the second place, we have 2DE = BF + FC = BC; or, 
this line DE, between the middle points of AB and AQ, is half - 
as long as BC. 

113. Cor.—If the lines AB and AC be made twice their 
present length, the line connecting them will be parallel to BC, 
and twice as long; and the new triangle will be equiangular to 
ABC; and we may form any number of new triangles by re- 
peatedly doubling the sides including A, which triangles will 


32 ELEMENTS OF GEOMETRY. 


have the third sides also doubled in the same way, and be equi- 
angular to ABC and to each other. 

114. Prop. 41.—J/f two lines lying m the same plane meet a 
third line, making the sum of the angles lying between the 
lines, and on the same side of the third line, less than two right 
angles, they will meet on that side of the third line, of suffi- 
ciently produced. 

Let AC, BD, be two lines in the same plane, lying so that the 
sum of the angles BAC and ABD is less than two right angles ; 
then they will meet if sufficiently produced. 


c D 


>) 


A B 


Draw two lines EF, EG, from some point, E, making an angle 
with each other equal to what the sum of BAC and ABD lacks 
of two right angles. Lay off on EF and EG distances EH, EK, 
equal to each other. Join HK. The sum of the angles EHK 
and EKH. is equal (87) to that of BAC and ABD; and it 
would be so wherever H and K might have been placed on the 
line, whether at equal distances from E or not. As they are 
at equal distances, the angles EHK and EKH are equal, and 
each is equal to (BAC + ABD), and consequently greater 
than BAC, which is supposed to be the smaller of the two. 

Now swing the line HK in the plane KHE round H (which 
corresponds with A in position) to the position HE. At the 
outset its angle with HE is, as just observed, greater than BAC. 
At the end it is less, being zero. 

Hence, at some intermediate position, represented by the 
dotted line, it must be equal to BAC; and of course, in this 
position the angle EKH will be equal to ABD. 

There are three cases in the remaining part of the demon- 
stration, according as HK is equal to, greater than, or less than 
AB. ’ 

First, if HK is equal to AB, we can superpose AB upon it 


ELEMENTS OF GEOMETRY. Se 


with the lines AC and DB attached; and (the planes being 
made to coincide) the line AC will fall on HE and BD on KE; 
they will therefore meet at E. 

Secondly, if HK is greater than AB, move it along the lines 
EF and EG toward EK, keeping it always at the same angle 
with E}’, and consequently always parallel to its previous po- 
sitions, and hence at the same angle with EG. There will, by 
the law of continuity, be some position in which it will be equal 
to AB, since when it arrives at E, it will be equal to zero; when 
it arrives at this position, superpose AB as before. 

Thirdly, if HI is less than AB, take EH and EK, two, four, 
eight, etc., times as long as they are now, according as HK re- 
quires to be multiplied by two, four, eight, ete., in order to 
equal or exceed AD. 

This being done, HK will (118) equal or exceed AB. If it 
equals Ab, superpose AB at once; if it exceeds AB, move it 
back toward E till it equals it, and superpose AB. 

In any case, then, we shall be able to superpose AC and BD 
on two lines that meet; they will therefore themselves meet if 
sufficiently produced. 

115. Cor.—Any two lines, lying in the same plane, and not 
parallel, will meet if sufficiently produced; or any two lines 
which lie in the same plane and do not meet, are parallel. 

116. Prop. 42.—J/f any two lines, lying in the same plane, 
be connected by three equidistant parallels, the middle one of 
these parallels will be equal to half the sum of the other two. 

Let the two lines LM, NO be connected by three equidistant 
parallels, AB, CD, EF. If the lines 
LM and NO are parallel, ABDC and is 
CDFE are parallelograms (111); hence, a 
AB, CD, and EF are all equal, and the 
truth of the proposition is evident. 

If they are not parallel, they will 
(115) meet in some point, K. Let AB 
be the nearest of the three parallels to 
K. Draw AF. The sum of the angles 
BAE, ABF is greater than two right angles; the sum of BAF 
and ABF is less than two right angles; hence for some point, 
Hon EF, between E and F, the sum of BAH and ABI will 

3 : 





34 ELEMENTS OF GEOMETRY. 


be equal to two right angles, and the lines AH, BF will (100) 
be parallel. Let G be the point where AH cuts CD. 

AB, GD, and HF are equal (111 and 106). And if we drop 
perpendiculars GP, GQ on AB and EF, the triangles APG, 
HiQG will be equal (89 and 51), the angles PAG and GHQ 
being equal. Hence, AG = GH; and similarly it may be 
shown that AC = CE. Hence (112), CG = 4EH; hence, CD, 
which is equal to CG + GD, is equal to (EH + 2GD), or to 
4(EH + HF + AB), or to 4(EF + AB). 

117. Cor. 1—EF — CD = CD — AB, since EF + AB = 
2CD; and if another parallel were drawn below EF, it would 
exceed EF as much as EF exceeds CD, or as CD exceeds AB. 
Or, if any number of equidistant parallels be drawn connecting 
two diverging lines in the same plane, any one will be longer 
than any other one which is nearer the point of meeting of 
the lines; and the difference of length between any two con- 
secutive ones will be always the same. 

118. Cor. 2.—The difference between any two of a set of 
equidistant parallels connecting two lines will be the same 
multiple of the difference between two adjacent ones, that the 
distance between the two parallels is of the uniform distance 
between two adjacent ones. 

119. Cor. 8.—The lines AG and GH. being equal, by the 
equality of the triangles APG and HQG, it follows that any 
line cutting a set of equidistant parallels is divided into equal 
parts by them. 

120. Cor. 4.—Hence, the distance between the parallels, in 

, Cor. 2, may be measured by any line 
cutting them, instead of on a perpen- 
dicular. 

121. Prov. 48.—The difference be- 
tween any two parallels connecting two 
lines is to the difference between any 
other two parallels connecting the same 
lines, as the distance between the first 
two parallels, measured on any line, as 
to the distance between the second two measured on the same 
line. : 

Let AB, CD, EF, GH. be four parallels connecting the lines 





ELEMENTS OF GEOMETRY. 35. 


AG and BH, and let LO be any line cutting them; we have 
GH—EF NO 
CD -SaB LM 

If LM and NO are commensurable, the truth of the proposition 
is evident ; for a set of parallels may be drawn between AB and 
CD, and between EF and GH, at a distance from each other, 
measured on LO, equal to the common unit of measure of LM 
and NO; and CD — AB will (118) be the same multiple of the 
difference between two adjacent parallels of this set that LM is of 
the distance measured on LO between these two. The same will 
GH — EF 
be true of GH — EF and NO. We shall then have GD AB 


and a both equal to the ratio of one of these multiples to the 


other, and hence, equal to each other. 

If LM and NO are not commensurable, suppose the propo- 
GH—EF NP 
CD—AB” LW 

Divide LM into equal parts, each smaller than PO; and lay 
off parts equal to these from N to O. At least one point of 
division, Q, between the parts will fall between P and O. Let 
RS be a parallel drawn through Q. We then have, by what 
has just been shown, since LM and NQ are commensurable, 


RS—EF NQ. | 
Se ap LM? but (117) GH > RS; hence, we shall have 


to show that 


sition false, and that 


oi EKEF NQ Ni eon Te TAN, 
fen AD eS me and hence, —— LM Bes Lit? but this is absurd, 
since NP < NQ; and asimilar absurdity would result if the 
GH — EF 


is neither 





point P were on NO prolonged ; hence, -5——a7 


less nor greater than, but precisely equal to ae 
122. Cor. 1.—The proposition is equally true if two of the 
GH—CD_ MO 
parallels coincide ; we have, for instance, CDs BoM: 
123. Cor. 2.—It is also true, if one of the parallels passes 
through K, the point of meeting of the lines, in which case we 
GH—COD _IT. GH jit MGs b 
Cle KL a CDa li Kla a Bike 





have, for instance, 


36 ELEMENTS OF GEOMETRY. 


or the length of any parallel is the length of any other, as its 
distance from the point of meeting of the lines, measured on 
any line, is to the distance of the other from the same point, 
measured on the same line. 

124. Cor. 3.—It is evident that the ratio of the distances 
between the parallels measured on different lines cutting them 
are the same, each being equal to the ratio of the differences of 

NO) 7 GEG? <2 EE 


the parallels. Thus LM AG > BD "2° each is equal to 
GH — EF 


6D — AB 

125. Cor. 4.—We have in the triangles KCD, KGH, by 
Cor. 2, ae = ee aa and moreover, the angles opposite 
GH, KG, and KH are equal to those opposite CD, KC, and 
KD; in the first case because they are identical, and in the 
other two cases by Prop. 386. 

126. Cor. 5.—If in a triangle, KGH, we lay off KC and 
KD proportional to KG and KH, so that Ke the line 

KG KH’ 
CD will be parallel to GH. 

For if we draw a line from C parallel to GH, it will, by 
Cor. 8, cut KH ata point distant from K by ce x KH; but 
this is equal to KD; CD therefore is this parallel. 

12 Cet Os. —The method of demonstration employed in 
this proposition may be employed to prove the following 
proposition, of which the above is a particular case ; paniele 
that «f a quantity in any way depending on another (as the 
difference of the two parallels depends on their distance) 
changes by equal amounts for equal changes of the other (as in 
this case, by Prop. 42, Cor. 1), whatever the amount of these 
changes may be, then will any two of tts changes, whether 
commensurable or not, have the same ratio to each other that the 
corresponding changes of the other quantity have to each other. 

128. Defi—Triangles, like KCD and KGH, having equal 
angles, and the sides opposite these equal angles proportional, 
are said to be semalar. The sides opposite the equal angles are 
called corresponding sides. 


ELEMENTS OF GEOMETRY. 3f 


129. Def:—Triangles are said to be similar though the equal 
angles do not lie in the same order as we go round the triangles 
from left to right, or from right to left, for the angles can be 
made to lie in the same order by the reversal of one of the tri- 
angles on its other side. 

We proceed to demonstrate several propositions with regard 
to similar triangles resembling those which have already been 
proved for equal triangles, which it is evident are a particular 
class of similar triangles. 

130. Prop. 44.—A triangle having two angles equal to 
two angles of another triangle, is similar to that other trt- 
angle. 

In the triangle DEF let the angles j : 
D and E be equal to A and B respec- 
tively in the triangle ABC; then will va 
DEF be similar to ABC. aie rat meee 

Let AB be the greater of the two 
sides included between the equal angles. 

Lay off AG = DE, and draw GH paral- 8 C 

lel to BC. The Htensle AGH will be 

similar (125) to ABC ; but it is equal (51) to DEF; hence 
DEF is similar to ABC. 

131. Prop. 45.—A triangle which has one angle equal to one 
angle of another triangle, and the sides including that angle 
proportional to those including the equal angle in the other 
triangle, 1s similar to that other triangle. 

In the triangle DEF’, using the figure of the last proposition, 
let the angle D be equal to the angle A of the triangle ABO; 
and let ae aa , AB and AC being longer than DE and DF. 

Lay off on AB/and AC, AG and AH equal to DE and DF; 
the triangle AGH will be similar (126) to ABC; but it is 
equal (28) to DEF; hence, DEF is similar to ABC. 

132. Prov. 46.—A triangle which has tts three sides pro- 
portional to the three sides of another triangle, rs similar to 
that other triangle. 

In the triangles ABO and DEF, using the figure of Prop. 

fee BO. AC 


44, let — DEER DF: then will the triangles be similar. 


38 ELEMENTS OF GEOMETRY. 


AB, BC, and AC being supposed longer than DE, EF, and 
DF; lay off on AB and AC, AG = DE and AH = DF. The 
triangle AGH will be similar (126) to ABC; and hence, we 
shall have je aoe = ee Hence, since AG and AH are 
equal to DE and DF, we shall have GH = EF. Hence, the 
triangles AGH and DEF are equal (48); and hence, DEF is 
similar to ABC. 

133. Prope. 47.—/f two points are moved equal distances in 
parallel lines, the line connecting them will be parallel to tts 
previous position, and will make the same angle with any fixed 
line that it did before the motion. 

Let the line connecting the points A and B meet the line CD 
in C; and let these points now be moved 
equal distances, AA’, BB’, in parallel 
lines; the line A’B’ will make with 
CD and angle, A’C’D, equal to the angle 
ACD. 

For the figure ABB’A’ is a parallelo- 
gram, by definition ; hence, the sides AB 
and A’B’ are parallel; hence, the angles 
ACD and A’O’D are (99) equal. 

134. Cor.—lIf all the points of any straight line be moved 
the same distance in parallel lines, they will in their new po- 
sitions also form a straight line. 

135. Prov. 48.—/f any three points are moved equal dis- 
tances in parallel lines, the triangle which they form after the 
motion will be equal to the triangle which they formed 
before. 

Let the three points A, B, C, be moved equal distances, A.A’, 

BB’, CC’, in parallel lines; then will 
; a the triangle A’B’C’ be equal to the 
A Lo Te e triangle ABC. 
For ABB’A’, ACC’A’, and BCO’B’ 
are parallelograms, as in the last pro- 
position ; hence, the lines A’B’, A/C’, 
y Pee TO ee 7 and B’C’ are equal to AB, AC, and 
BC, respectively ; hence, the triangle 
A’B’C’ is (48) equal to the triangle ABC. 





ELEMENTS OF GEOMETRY. . 39 


136. Schol.—The proposition is equally true, whether the 
parallels lie in the plane of the first triangle or not. If they 
do, the second triangle is also in the same plane. 

137. Def—The motion treated of in these two last propo- 
sitions, In which a number of points are moved equal distances 
in parallel lines, is called translation. 

138. Prop. 49.—An angle, the sides of which are equally 
inclined to the sides of another angle in the same plane, is 
equal to that other angle. 

Let the sides B’A’, B’C’, of the angle A’B’C’ be equally in- 
clined to the sides BA, BC of the angle 
ABC; that is, let the angle A’DA be ras 
equal to the angle C’EC; then will the ai 
angle A’B’C’ be equal to the angle 
ABC. 

For if we move the point B’ to B, and 
at the same time move A’ and C’ dis- 
tances A’A” and O’O” equal to B’B in 
lines parallel to B’B, the angles which 
B’/A’‘ and B’C’ make in their new positions, BA” and BC”, with 
BA and BC, will (183) remain unchanged; hence, the angles 
A” BA and C”’BC are equal, since A’DA and C’EC were equal ; 
hence, A” BC” and ABC are equal, since A’-BC” = A” BC — 
C’'BC, and ABC = A” BOC — A” BA. | 

But A” BC” and A’B’O’ are (135) equal; hence, A’B’C’ and 

ABC are equal. 
' 1389. Cor.—aA triangle, two angles of which have their sides 
equally inclined to those of two angles of another triangle, is 
similar to that other triangle. 

140. Schol. 1.—In this proposition and its corollary the 
angles or inclinations of the sides of any angle to those of any 
other must both be counted in the same way, that is, both from 
left to right, or both from right to left. This conditions how- 
ever, is unnecessary when the sides of one angle are parallel or 
perpendicular to those of another; in this case, no ambiguity 
can arise. 

141. Schol. 2.—The angle A’BC’ may be made to coincide 
with ABC by first moving it to the position A’”-BC” by a motion 
of translation, and then rotating it around an axis of its plane 





40 ELEMENTS OF GEOMETRY. 


passing through B. We shall show hereafter that any angle 
in space may similarly be made to coincide with any other 
equal angle (either side of the first being made to coincide 
with either side of the second), by one translation and one 
rotation. 

142. Prop. 50.—A polygon can be constructed, having its 
angles equal to those of any gwen plane polygon, taken in the 
same order, and having its sides in any required uniform ratio 
to the sides between the equal angles in the gwen polygon. 

Let ABCDE be the given polygon. Draw from some point, 
H, within, the ines HA, HB, HC, HD, 
HE, and lay off on these lines (prolonged 
as in the figure if the sides of the new 
polygon are to be longer than those of the 
given one) HA’, HB’, HO ap ae 
equal to 7. TA, r. HB yi Bee 
yr. HE respectively, 7 being the uniform 
ratio required. Draw A’B’, B/C’, C’D’, 
D’'E’; A’B’O'D’E’ will be the required polygon. 

For the triangles HA’B’ and HB’C’ are similar (181) to 
HAB and HBC respectively; hence, the angles HB’A’ and 
HB’C’ are equal to HBA and HBC respectively, and hence 
their sums, A’B’C’ and ABC, are equal; and the same could be 
proved for the other angles of the two polygons. 

Also, by the similarity of the triangles, each of the sides of 
the new polygon is 7 times the side parallel to it in the given 
polygon ; hence, the new polygon has the properties required. 

143. Def—Polygons thus related are said to be simelar. 
There is no need that either of them should he’in any one 
plane; though the demonstration given above only applies to 
plane polygons, since otherwise A’B’O' and ABC are not the 
sums of HB’A’ and HB’C’, and WBA and HBC. It is only of 
these that we shall have occasion to speak. 

144. The sides having the uniform ratio to each other are 
called corresponding sides, as in the case of similar triangles. 
Two points, F and F’, situated on the corresponding sides in 
BF DEY 2 By ies 
BCU BC TBF RBG 
corresponding points, and lines such as FG and F’G’, connecting 





such a way that == 7, are. called 


ELEMENTS OF GEOMETRY. 41 


two pairs of corresponding points, corresponding lines. It is 
/ / / 
plain that ar = ae = r, and hence that oe also = 1. 

145. Def:—The sum of the sides of a triangle or polygon is 
called its perimeter. The perimeters of two faiths triangles 
or polygons are evidently in the same ratio 7; and the same is 
true of the portions of the perimeters included between cor- 
responding points. 

146. Schol——Any number of new similar polygons, having 
the same uniform ratio of the sides, though not the same angles, 
as the original polygons, may evidently be formed by means of 
corresponding lines. 

147. Prop. 51.—A line bisecting any angle of a triangle, 
divides the opposite side into parts proportional to the adjacent 
sides. 

Let BD bisect the angle ABC; then 


E 
Peon bo 
will Maat AB’ For prolong AB to E, RP 
making BE = BC, and draw CE. The 
triangle CBE is isosceles; hence (88), the 4 > C 


angle ABC is equal to twice the angle 
BCE. Hence, DBC = BCE. Hence (99), DB and CE are paral- 
lel ; hence (124), <= a But BE = BC; hence, a =< an 
148. Prov. 52.—T/f, in a right-angled triangle, a perpen- 
dicular be dropped from the right angle on the hypothenuse ; 
1. The two triangles thus formed will be similar to the 
original one, and hence to each other. 
2. Hach side about the right angle B 
in the original triangle will be a mean 
proportional between the whole hypothe- 
nuse and that part of tt which is adja- amr 
cent to the side , 
8. The perpendicular will be a mean proportional between 
the parts of the hypothenuse. 
Let ABC be a triangle, right-angled at B, and let BD be 
perpendicular to AC. 
1. The triangles ABC, ABD, are (180) similar; the same is 
true of ABC and CDB. 


49 - ELEMENTS OF GEOMETRY. 


AB AC OB AC 
mens 9 Np) ig MEA cL = (OES 


3. Hence, from the similarity of ADB and CDB 2 


RD a, Olas 
149. Cor. 1.—Hence, from the second part of the proposition, 
CB TA Bia AC se see AD CBG. 
’OD AD” OB” AB’ © OD AB ae 
JM BYE ee sy , 
CD ~ CB 

150. Cor. 2.—Tlence, also, from the second part, we have 
AB’? = AD. AOC, and CB? = CD.AQ; adding, we have, since 
AD + GD SBOrAG ee Ub awe 

151. Schol——By. AB’, Cb’, and AC’, in these corollaries, is 
meant the algebraical square or second power of the lines AB, 
CB, and AC, expressed in terms of any definite unit of measure. 
The geometrical meaning of the equations will be seen in the 
part of the subject upon which we now enter. 


we have 





ISO (OA Te 1 Ae 


AREAS OF PLANE FIGURES. 


152. Def—A line either broken or curved, or partly broken 
and partly curved, lying in a plane and returning into itself, is 
called a plane figure. If it lies on a curved surface, it takes its 
designation similarly from the character of that surface. 

1538. Def—The portion of the surface enclosed by the figure 
is called the area of the figure. It may be conceived, as stated 
(7), to be the result of the movement of aline. The areas of two 
plane figures are evidently equal when one may be superposed 
on the other by a different arrangement of its parts. 

Also, the sums and differences of equal areas are equal. 

154. Def—Any side of a triangle being called its dase, the 
perpendicular from the opposite angle on the base, or the base 
prolonged, is called its altitude ; and any side of a parallelo- 
gram being called its base, the perpendicular between it and the 
opposite side is called its ad¢itude. A triangle has three bases 
and altitudes; a parallelogram has four bases and two altitudes. 

155. Prop. 53.—Zwo parallelograms, having equal bases 
and equal altitudes, wre equal wr area. 

The bases of two such parallelograms may be superposed ; and 
since their altitudes are equal, their op- 
posite bases will fall in the same line,as ¢ ¢£ fo 
in the figure, in which AB represents the 
coinciding bases, and CD and EF the op- 
posite bases. 

Now CD = EF; hence CF —CD= A B 
CF — EF, or DF = CE; also BD = AO, 
and BF = AE; hence the triangles CAE and DBF are (48) 
equal; and hence the parallelograms are equal in area, since 
they are formed by a different arrangement of the quadrilateral 


44 ELEMENTS OF GEOMETRY. 


AEDB, and these equal triangles, or by the addition of 
equal triangles to AEDB, or by the subtraction of equal tri- 
angles from ACI'B. 

156. Prop. 54.—TZhe areas of any two rectangles are to each 
other as the products of their bases and altitudes ; these bases 
and altitudes being expressed numerically m terms of any one 
unit of measure. 

Let ABDC, EFHG, be two rectangles, AC, EG being their 

altitudes. Draw in the one having the 
4 greater altitude a parallel, KL, to the 
of 0 LE r _ bases, at a distance KC from CD one of 
the bases, equal to the altitude of the 
other rectangle. 

The new rectangle, KLDC, or KD, as 
cw D GH it may be called, may be shown (127) 

to be to AD as KC is to AC, whether 
KC is commensurable to AC or not; for if a line moves from 
CD to AB, keeping always parallel to its previous positions, 
the difference between any two rectangles which it forms with 
CD, and the lines described by its extremities, will be always 
the same for the same amount of movement of the line. 

In the same way, drawing MN between, and perpendicular 
to KL and CD, at a distance from KOC equal to EF’, we have 


UY ye CNG) AD AC 
KN > GN? but Kp * has been shown, = KG} hence, multi- 
AD CDE Aw 


plying these equations together, KN > ON KO 
And if we take for CD, CN, AC, and KC their numerical 
values in terms of the unit of measure, this will still hold true; 


hence, using CD, etc., in this sense, we have od = OO ae 
KN CN.KOC 
which was to be proved. 

157. Cor. 1—If CN and KO are equal to each other and to 
the unit of linear measure, KN will be a square, the side of 
which is this unit of measure; and if such a square be taken 
as the unit of measure for surfaces, or the unit of superficial 
measure, as it is called, we have AD = CD. AC; or the num- 
ber of superficial units in a rectangle equal to the product of 
the number of linear units in its base multiplied by the number 


ELEMENTS OF GEOMETRY. 45 


of linear units in its altitude; or, as it may be concisely ex- 
pressed, the area of a rectangle 1s equal to the product of 
its base and altitude. 

158. Cor. 2,—The same is true (155) of any parallelogram. 

159. Cor. 3.—If to any triangle ACD (see figure of Prop. 38) 
an equal one, DBA, be added, a parallelogram will be formed ; 
hence this parallelogram will have double the area of the tri- 
angle; but if any base of the parallelogram be taken as the 
base of one of the triangles, this triangle will have the same 
altitude as the parallelogram; hence, we may say in the same 
sense, as in the case of the parallelogram, that the area of a tri- 
angle is equal to half the product of its base and altitude; and 
it is equal to half that of any parallelogram having the same 
base and altitude. 

It is evidently immaterial which side of any triangle be taken 
as the base in this expression for the area, if the perpendicular 
on that side from the opposite angle be taken as the altitude. 

160. Cor. 4.—Parallelograms and triangles which have the 
game base are to each other in area as their altitudes; if they 
have the same altitude, they are to each other as their bases. 

161. Def—A. Trapezoid is a quadrilateral, two opposite 
sides of which are parallel, but unequal. The other two sides 
are not parallel, for if they were, the figure would (111) be a 
parallelogram, and have both pairs of opposite sides equal. 

The two sides which are not parallel will meet if produced 
beyond the shorter of the two parallel sides, forming thereby a 
triangle with each of the parallels. 

162. Def—tThe parallel sides are called the dases of the 
trapezoid, and the distance between them, its altetude. 

168. Prop. 55.—The area of a trapezoid is equal to half 
the product of the sum of its bases, and its altitude. 

Let ABCD be a trapezoid, AB and DC 
being parallel, but not equal. Divide it 4 B 
into two triangles by the diagonal AC. 
Taking AB and CD for the bases of these 
triangles, they have a common altitude, 
which is the altitude of the trapezoid. 
Their areas are equal to this altitude multiplied by $A B and 
4CD respectively. Hence the sunf of their areas, or the area 


E, V2 


D C 


A6 ELEMENTS OF GEOMETRY. 


of the trapezoid, is equal to this altitude multiplied by 
4(AB + CD); which was to be proved. 

164. Cor. 1.—If we draw a line EF connecting AD and BC, 
parallel to AB and DC and equidistant from them, this line 
will (116) be equal to half the sum of AB and DC; so that the 
area of the trapezoid is equivalent to the product of a line 
drawn in it half way between the bases, and the altitude. 

165. Cor, 2.—Calling a line drawn parallel to the bases, and 
connecting the other sides, the breadth of the trapezoid, and 
drawing an odd and indefinitely great number of such breadths 
at indefinitely small but equal distances, it is evident (116) that 
the middle one will be the mean or average breadth of the 
trapezoid, being the half sum or arithmetical mean of any 
two at equal distances from it; hence, the area of a trapezoid 
is equal to its average breadth multiplied by its altitude. 
Using the word breadth in a similar sense, the same is true of 
a triangle, which, indeed, is only a trapezoid, one of whose bases 
is equal to zero; and the same is true of any plane figure; since 
it may be divided into an infinite number of trapezoids by 
parallel lines ; its average breadth being understood to be the 
average length of these parallels when they are equidistant ; 
and indeed it is true of any surface which can be formed by a 
straight line which keeps always parallel to its previous posi- 
tions, whether it constantly retains its length, or changes it in 
any way, perhaps breaking up into various parts; if the altitude 
in this case is the line, straight or curved, connecting all the po- 
sitions of the forming straight line and perpendicular to them 
all; or, in other words, if the altitude is the amount of move- 
ment of this forming line perpendicular to itself. 

166. Cor. 3.—Since a parallelogram is the limiting form of 
the trapezoid, the expression for the area of a parallelogram 
can be deduced from that of the trapezoid, by considering the 
bases of the trapezoid as equal. 

167. Def—By the square of a line is meant a square whose 
side is equal to that line. Its area in superficial units equal to 
the square of the linear unit, is equal, as has been shown (157), 
to the arithmetical square of the number of linear units in its 
side; and it may be represented in the usual algebraical way. 
Thus the area of the square Of the line AB is represented by AB?. 


ELEMENTS OF GEOMETRY. 47 


168. Def—By the rectangle of two lines is meant a rectangle 
whose base and altitude are equal to those lines respectively ; 
either side of the rectangle being taken for the base, the distance 
between it and the opposite one being the altitude. Its area, 
like that of the square, may be represented in the algebraical 
way, by the product of the two lines. Thus the area of the 
rectangle of the lines AB and BC may be represented by 
AB . BC; being really equal in superficial units, equivalent to 
the square of the linear unit, to the number of linear units in 
one line multiplied by the number in the other. 

169. Prop. 56.—The square of the sum of two lines is equal 
to the sum of the squares of the lines, increased by twice the 
rectangle of the lines. 

Let AB, BC be the two lines. Describe 


a square, ACDF, on their sum, AC. Lay wanes the 


off AG = AB, and draw GK, BE parallel 
to AC and AF respectively. AH is the 
square on AB. ED and HK are each 
equal to BC; hence EK is the square on 


BC. FH and HC are each rectangles on 
AB and BC. But AD is equal to AH + EK + FH + a 
hence (AB + BC)? = AB* + BO? + 2AB. BC. 

170. Prop. 57.— The square of the difference of two lines is 
equal to the sum of the squares of the lines, diminished by 
twice the rectangle of the lines. 

Using the same figure as in the last proposition, let AC and 
BC be the lines. The square on AB, their difference, is equal 
to the square on AO diminished by EK, FH, and HC; or 
it is equal to the square on AC increased by EK, and dimin 
ished by FH + EK, and EK + HC. But each of these last 
quantities is equal to the rectangle of AC and BC. Hence 
(AC — BC’ = AC? + BC? — 2AC. BC. 

171. Prov. 58.— The rectangle of the sum and difference of 
- two lines is equal to the difference of their squares. 

In the figure of Prop. 56, the difference of the squares of 
AC and BC is equalto AK + FH. But if FH. is placed along- 
side of AK so that FG will be the continuation of AC, the 
whole rectangle thus formed will be the rectangle of AC + BC 
and AC — BC; hence (AC + BC) (AC — BC) = AC? — BC? 


48 ELEMENTS OF GEOMETRY. 


172. Schol.—These propositions agree with the results which 
would be obtained by algebra. 

173. Prop. 59.—The square of the hypothenuse of a right- 
angled treangle rs equal to the sum of the squares of the other 
two sedes. 

This proposition might be considered sufficiently established 
by Prop. 52, Cor. 2; we proceed, however, 
to give a purely geometrical demonstration. 

Let the triangle be ABC. Construct on 
AC, its hypothenuse, the square ACED. 
Let the angle A be less than, or at most 
equal to, the angle ©. Drop from D, DF 
perpendicular to AB, and prolong AB; 
the prolongation will (114) cut CE in some 
point G. This point will lie (52) at E, if the angle A of the tri- 
angle is equal to the angle CO, that is, to half a right angle; it 
will lie between C and E if the angle A is less than this. Drop 
on the prolongation of AB the perpendicular EN. Drop DM 
perpendicular to NE produced. Lay off AH = OG, and drop 
HK perpendicular to DI. Lastly, drop AL perpendicular to 
iH. produced. 

The angles BGC and DHK are each equal (99) to DAF; 
and (89) DAF = R — BAC, and hence = ACB. And ACB 
= DEM (138), since NM sal CB are both perpendicular to 
AN, and therefore parallel. Also, the angles AHL and EGN 
are equal (26) to DOK and BGC respectively... Hence, the 
angles ACB, BGO, DHK, DEM, AHL, and EGN are all 
equal. Hence, the triangles CBG, DHK, and ABC are equal 
(61) to ALH, ENG, and DME respectively ; hence, the square 
ADEC is equal in area to the figure ALKDMNA, since the 
first is equal to the figure AHKDEGA increased by the first 
three triangles, and the second is equal to the same figure in- 
creased by the second three. 

But the figure ALKDMNA is composed of FDMN and 
ALKF; and both these (97) are rectangles, since three of the 
angles if each are right angles by construction. Furthermore, 
DM and DF are each equal to AB, and AL and AF to CB, by 
the equality of the triangles just shown, and the equality (61) 
of the triangle DFA to ABC; hence, FDMN and ALKF are 





ELEMENTS OF GEOMETRY. 49 


the squares of AB and CB respectively ; and ADEC, the square 
of AO, is equal to their sum, as was to be proved. 

The demonstration also holds when the prolongation of AB 
cuts CE at E; in this case the triangles DKH and ENG 
disappear. 

174. Prov. 60.—The square of a side opposite an acute 
or obtuse angle in any triangle, is equal to the sum of the 
squares of the other two sides, diminished in case of the acute 
angle, increased in the case of the obtuse, by twice the rectangle 
of either of the sides mcluding the angle, and the distance from. 
the vertex of the angle to the foot of the perpendicular dropped 
on that side or that side produced, from the angle opposite to it. 

In the triangle ABC, let A be the acute angle considered. 
Whether BD falls, as in the figure, on AC 


produced, or on AC itself, we should (170) £ 
have CD? = AC? + AD? — 2AC. AD. 

Adding BD? to both members, we have 
(173) BO? = AC? + AB*? — 2AC. AD, a oe 


which was to be proved. For the case 
of the obtuse angle C, we have (169) ae = BC’ Clee 
2AC .OD. 

Adding BD? to both members, as before, we have (173) AB? 
pts ie DC? + 2A0...CD. 

175. Cor.—lf the square of any side of a triangle is equal to 
the sum of the squares of the other two, the angle opposite that 
side isa right angle. Jor if it were acute or obtuse, the square 
of the side opposite it would be less or greater than the sum of 
the squares of the other two, by this proposition. 

176. Prov. 61.—/n any triangle, the sum of the squares of 
any two sides is equal to twice the sum of the squares of 
half the third side, and a line drawn to the middle point of 
the third side from the vertex of the opposite angle. 

In the triangle ABO, let E be the middle point of AC, and 
‘BD the perpendicular from B on AC. “ 


We have (174) BC? = EC? + EB? — 
2EC. EB, and AB? = EA? + EB? + 
2A .EB. But EA = EC; hence, . 7 


adding these two equations, we have a 
Bo + nae = 2(EC? + EB’), which was to |e proved. 


50 ELEMENTS OF GEOMETRY. 


177. Schol.—It would make no difference if BD fell on AC 
produced in either direction; one of the angles at E would be 
acute in any case, and the other obtuse. 

178. Cor.—The sum of the squares of the sides of a paral- 
lelogram is equivalent to the sum of the squares of the diago- 
nals, Let ABDO, using the figure of Prop. 38, be a parallelo- 
gram. The diagonals AD, BC bisect each other (107). Hence, 
by the proposition just proved, AB? + AC? = 2(EB? + EA?); 
also DB? + DC? = 2(EB* + EA’). Hence, AB* + AC? + DB? 
+ DO = 4(EB? + EA*) = (2QEB) + QEAY = BC? + AD? 
which was to be proved. 

179. Prov. 62.—The areas of two triangles, which have an 
angle in each the same, are to each other as the rectangles of 
the sides including the equal angles. 

Let ABC, DEF, be two triangles, having the angles A and 

area of ABC AB. AG s 
D equal. Then will oa of DER = DED DF Lay off AG 

Phe D and AH equal to DE and DF respec- 
tively, on AB and AQ, produced if ne- 
eessary. Draw GH. We have (160) 











‘s areaof AGC AC areaof ABC AB 
C 3 areacf AGH AH’ areaotAGO AG 
Multiplying these equations together, we 

B C area, of ABC AB. AC 
have or 





_area of AGH AG. AH’ 
eee eee aa aa since AG = DE, AH = DPF, and 


the triangles AGH and DEF are (28) equal. 

180. Cor—If ABC and DEF are similar, we shall have 
TAUB pees hence, 22 of AGC _ area of ABC | or themes 
AH 7 AGY -areaof AGH area of AGC’ 
of AGC will be a mean proportional between those of ABC 
and AGH or DEF. 

181. Prop. 63.— The areas of any two similar polygons are 
to each other as the squares of their corresponding sides. 

Using the figure of Prop. 50, we have (179), for the triangles 
HA, HA B., | 


Area of HA’B’ TIA’ . HB’ } 
Area of HAB” HA.HB mie’) 








ELEMENTS OF GEOMETRY. 51 


In the same way, 
Area of HB’O’ 
Area of HBC 
and so on for the rest. And since the separate triangles, of 
which the polygons are composed, are to each other in this ratio, 
the whole polygons are in the same ratio; which is the square 
of the ratio of the separate sides or of the whole perimeters to 
each other. | 
182. Cor.—The areas of any two similar polygons are to 
each other (144) as the squares of any corresponding lines, — 
183. Schol.—Triangles, being polygons, are included in the 
above demonstrations. 


Vip 


a CO ays 


THE CIRCLE. 


184. Def—The curve described, or formed, by any point on 
a forming line of a plane, is called a circle. 

185. Def-—The centre of the plane around which the form: 
ing line revolves is also called the centre of the circle. 

186. Def—The distance on the forming line between the 
centre and the forming point is called the radius (plural radi) 
of the circle. 

187. Def—A line connecting any two points of the circle is 
called a chord, and the portion of the circle between these 
points is called an arc. Any chord of course has two ares cor- 
responding to it, one on one side, the other on the other. The 
chord is said to swhtend the arcs, and the arcs to be subtended 
by the chord. When the two points are opposite, so that the 
chord passes through the centre, the chord is called the dvameter 
of the circle, and is equal to twice the radius; and the two ares, 
which are manifestly equal, since they must admit of superpo- 
sition, by revolving one of them round the diameter, or be in 
some points unequally distant from the centre, are called semz- 
circles, such being equal to half the whole circle. Half a semi- 
circle, or one quarter of a circle, is called a guadrant. 

188. Defi—A_ segment of a circle is the figure formed by a 
chord and the are which it subtends. 

189. Def—A. sector of a circle is the figure formed by two 
radii and the are connecting them. 

190. Defi—By the angle corresponding to an are, or to a 
chord, is meant the angle made at the centre by the radii drawn 
to the extremities of the are or chord. Thearcor chord is also 
said to correspond to the angle. 

191. Def—KEqual circles are those which have equal radii ;. 
if their centres are superposed, they will coincide. 





ELEMENTS OF GEOMETRY. 53 


192. Prop. 64.—Jn the same circle, or in equal circles, any 
two ares are to each other as the corresponding angles, and 
VICE VErSA. 

Superpose the circles if there are two. 

In the first place, if the angles are equal, the ares are equal ; 
for the equal angles, if they do not coincide, may be made to 
do so by rotation; and when they coincide, the extremities of 
the radii forming them will coincide, since these radii are 
equal; and, as the circles coincide throughout, the arcs con- 
necting these extremities will coincide, and be equal. 

Similarly, if the arcs are equal, the angles are equal; for 
the equal arcs can be made to coincide, and then the sides 
of the angles will coincide, and the angles therefore will be 
equal. 

Hence, in the same circle or in equal circles, for every equal 
change in any angle, there is an equal change in the corre- 
sponding are, and vce versa. 

Then (127) any two ares will be to each other as the corre- 
sponding angles, and vice versa; which was to be proved. 
198. Schol.—The angle at the centre may be said to be 
measured by the arc; that is to say, it is the same part of four 
right angles that the corresponding arc is of the circumference. 

194. Prov. 65.—Jn the same circles, or im equal circles, the 
nearer an angle rs to two right angles, the greater vs its chord ; 
and conversely, of two unequal chords, the greater has a cor- 
responding angle nearer to two right angles than the corre- 
sponding angle of the other. 

For the chord forms a triangle with the radii to its extremi- 
ties; hence, if one of two angles at the centre is nearer to two 
right angles than another, the chord opposite is greater (46) than 
the chord opposite to the other. And if two chords are unequal, 
the angle opposite the greater one is nearer to two right angles 
than the angle opposite the other, by the corollary of the same 
proposition. | 

195. Cor. 1.—Since the angles are (192) proportional to the, 
arcs, we may substitute arc for angle, and a semicircle for two 
right angles, in the above proposition. 

196. Cor. 2—A diameter is greater than any chord, as 
would also be evident by Prop. 14. , 


54 ELEMENTS OF GEOMETRY. 


197; Cor. 8.—Equal arcs are subtended by equal chords; 
since if the chords were unequal the arcs would be unequal. 
And vice versa, equal chords subtend equal arcs; if the chords 
are equal, the arcs are equal. 

- 198. Prov. 66.—The radius which bisects a chord is per- 
pendicular to the chord, and also bisects the are subtended by 
the chord. : 

Let AB bea chord of a circle, the centre of which is C. 
Draw CA and CB. 

CAB is an isosceles triangle; hence, if 
a radius, CD, be drawn, bisecting the chord 
at E, the angles at E will be right angles, 
and the angle ACB will (32) be bisected. 
But equal angles correspond (192) to equal 
arcs; hence the angles ACD and BCD 

D being equal, the arcs AD and BD are 
equal. . 

199. Cor. 1.—Only one perpendicular can be drawn (87) 
from the centre to the chord. Hence, any radius which is per- 
pendicular to a chord bisects it and its subtended are. Also, 
the perpendicular from the middle point of an arc on the sub- 
tending chord passes through the centre of the circle. 

200. Cor. 2.—The perpendicular in the plane of the circle 
to the chord at its middle point, or centre, passes through the 
centres of the circle and of the are. 

201. Cor. 3.—The centres of the circle, the chord, and its 
are lie in the same straight line; but two points suffice to de- 
termine a line; hence, any line which passes through two. of 
these centres passes through the third also. 

202. Prop. 67.—Jln the same circle, or in equal circles, the 
greater of the two chords is the nearer to the centre ; and con- 
versely, the nearer of the two chords to the centre is the greater. 

Superpose the circles, if there are two, and let AB and DE 
pe the chords, AB being the greater. Drop perpendiculars 
CF and CG on the chords from C, the centre of the circle or 
‘superposed circles. These perpendiculars bisect the chords 
(199). Draw CA and CD. | 

OF? + AK’ — OA? (173); and CG? + DG" == OD siya 
same; but CA = CD; hence, CF? + AF? = CG? + DG’ 


S 
& 


ELEMENTS OF GEOMETRY. 5D 


But AF is half of AB, and DG half of DE, as has been shown ; 
hence, since AB > DE by the hypothe- 

sis, AF > DG. Hence, CF < CG; which 2. 

was to be proved. 

Conversely, the nearer of the two 
chords to the centre is the greater; for 
nor < CG, AF > DG, since CF? + 
BY = :OGF-e DG’, 

208. Cor.—Chords equally distant 
from the centre are equal; for, if not, one of them would, by 
the first part of this proposition, be nearer to the centre than 
the other. And, conversely, equal chords are equally distant 
from the centre; for if not, they would be unequal, by the 
second part of the proposition. 

204. Prov. 68.—Through three points, not in the same 
straight line, one circle may be made to pass, and only one. 

Let A, B, C be three points not in the same straight line. 
The triangle ABC must (40) have two 
acute angles. Let D, Ebe the middle 
points of two sides opposite to acute 
angles. Draw DE, and erect perpen- 
diculars in the plane of ABC at D and 
E to AB and BC on that side on which 
AC lies. The angles BDE, BED will 
be equal (181) to A and C respectively ; 
hence they will be acute; hence DE will lie between the per- 
pendiculars DF and EG. The sum of the angles EDF and 
DEG will be less than that of BDF and BEG by the sum of 
BDE and BED; hence, it will be less than two right angles, 
since BDF and BEG are right angles. Hence, DF and EG 
will meet (114) at some point, H. 

Now this point, H, is equally distant (84) from A and B and 
from B and C; and it is in the plane of ABC, since both per- 
pendiculars have been drawn in that plane; hence a circle may 
be described around it with a radius IB, passing through A, 
B, and ©. 

Furthermore, the only points in the plane of ABO which 
are equally distant from A and B lie (34) in the line DH, 
and the only points in that plane equally distant from A 





56 ELEMENTS OF GEOMETRY. 


and © liz in the line EH. Hence, no circle can be drawn 
. through these points except that around H with a radius 
HB. 

205. Cor. 1.—Two circles cannot intersect in more than two 
points. 

206. Cor. 2.--Through three points, A, B, C, on the same 
straight line, no circle can be made to pass; for the perpen- 
diculars DF and EG would (71) be parallel, and would not 
meet in any common point, H. Or, in other words, a straight 
line cannot meet a circle in more than two points. 

207. Prov. 69.—Jf the distance of a line lying im the plane 
of any circle from the centre of that circle is greater than the 
radius of that circle, it will not meet it ut all ; if its distance 
is equal to the radius of the circle, it meets the circle only in 
one point ; if its distance is less than the radius of the circle, 
it mects it in two pornts. 

The first part of the proposition is evident from the fact 
that all points of the line are (41) at a distance from the 
centre greater than the radius. The second part is equally 
evident from the fact that all points of the line except 
the foot of the perpendicular on it from the centre are at 

a distance greater than the radius, 
yee Oe ea while the foot of this perpendicular 
eH kK, is at a distance equal to the radius. 
“ADB is such a line, CD being the 
perpendicular. 

But if a line, EF, lie at a distance 
from the centre less than the radius, 
denoting the foot of the perpendicular 
dropped from the centre on EF by G, 

we will lay off on each side of G, on the line EF, a distance 
the square of which shall be equal to CD? — CG*. Let the 
points thus determined be H and K. We shall have (173) 
CG? + GH’? = CH?, or GH? = CH? — CG*. But GH? has been 
constructed equal to CD*? — CG*; hence CH = OD. Similarly 
it may be shown that CK = CD. Hence H and K are on the 
circle. 

208. Cor.—Conversely, if a line does not meet a circle at all 
in the plane of which it lies, its. distance from the centre of the 


ELEMENTS OF GEOMETRY. 57 


circle is greater than the radius, for if it were equal to, or less 
than the radius, it would meet the circle at one or two points. 
The converse of the other two parts of the proposition may be 
similarly proved. 

209. Schol.—The second part of the proposition, and its con- 
verse, may be thus stated: A line which is perpendicular to a 
radius at its extremity, meets the circle at no point except the 
extremity of that radius; and a line which meets the circle 
only at one point, is perpendicular to the radius drawn to that 
point. 

210. Def—A line meeting or touching the circle only at one 
point is called a tangent to the circle ; the point where it meets 
it is called the point of contact. A line meeting a circle at 
two points is called a secant. A tangent may be regarded as a 
secant in which the two points are indefinitely near to each 
other. | 

211. Prop. 70.—TJf the distance between the centres of two 
circles which lie in the same plane is greater than the sum, or 
less than the difference of the radii, the circles do not meet each 
other at all ; of it 1s equal to the sum or to the difference of 
the radi, they meet each other only in one point ; if i 28 less 
than the sum and greater than the difference of the radia, they 
meet each other in two points. 

Let A and B be two centres. In the first case it is evident 
that no pointon the line AB, or AB 
produced in either direction, can be 
common to both circles ; for if a point 
Con AB were common to both, AB 
would be equal to the sum of the 
radii; andif a point D on AB pro- 
duced were common to both, AB 
would be equal to the difference of 
the radii. Nor can a point, such as E or I’, not on the line AB, 
be common to both circles; for, if it were, we should have in 
the triangle AEB or AFB, AB greater than the sum, or less 
than the difference of the other two sides; which is (42) im- 
possible. 

In the second case, if AB is equal to the sum of the radii, 
there will be a point, C, on the line AB common to both circles, 





58 ELEMENTS OF GEOMETRY. 


but no point on AB produced in either direction ; for if there 
were, AB would be equal to the difference of the radii as well 
as the sum, which is impossible. Nor. will there be any point, 
such as E, outside of AB common to both; for if there were, 
we should have in the triangle AEB, AB equal to the sum of 
the other two sides, which is (42) impossible. And if AB is 
equal to the difference of the radii, there will be a point on AB 
produced on the side of the smaller circle which will be com- 
mon to both, and no point on AB; for if there were, AB would 
be equal also to the sum of the radii, which is impossible; nor 
on AB produced in the other direction ; for if there were, the 
lesser radins would be greater than the greater one. Nor will 
there be any point, such as I’, outside of AB common to both 


circles ; for if there were, we should have in the triangle AFB, 


AB equal to the difference of the other two sides, which is (42) 
impossible. 

In the third case, a triangle can be constructed (48) with the 
radii and AB for sides. This triangle may be placed so as to 


have the side which is equal to AB on AB, and with the sides 


equal to the radii terminating at their proper centres; and this 
in two positions on opposite sides of AB, as AJiB and AGB, 
or AFB and AHB. The angle in this triangle opposite AB 
with in each of these positions be on both circles ; or the circles 
will meet in the two points E and G, or I and H, which it 
occupies. | 

212. Cor. 1.—Conversely, if two circles in the same plane 
do not meet, the distance between their centres must be greater 
than the sum, or less than the difference of the radii; for, if it 
were not, the circles would meet in one point, or two points. 
The converse of the other two parts of the proposition may be 
similarly proved. 


213. Cor. 2.—In the second case, the point of meeting of 


the circles is on the line connecting their centres. In the third 


case, the line connecting the ceutres is perpendicular to the 


line connecting the two points of meeting, and bisects it, by 
Prop. 10. | 

214. Schol.—lf the distance between the centres of two 
circles in the same plane is greater than, or equal to, the sum 
of the radii, the circles are outside of each other; if. it is less 


Oe 


nny ee ee 


ELEMENTS OF GEOMETRY, 59 


than the sum of the radii and greater than the difference, they 
are partly outside, partly inside, of each other ; if it is equal to, 
or less than the difference, the smaller is inside of the larger. 

215. Def-—Two circles which have one point of meeting are 
said to be tangent to each other; externally or imternally, 
according as the distance of the centres is equal to the sum, or 
to the difference of the radii. The point where they meet is 
called the point of contact. ‘Two circles which have two points 
of meeting are said to ¢ntersect each other. If the centres of 
two circles coincide they are said to be concentric. 

216. Defi—An inscribed angle is one which is formed by 
two chords meeting at a point on the circle. It is said to be 
inscribed in that part of the circle which is not included be- 
tween its sides. The angle formed by a tangent and a chord 
which meets it at the point of contact, may be considered as an 
inscribed angle, since the tangent may be supposed to pass 
through two points indefinitely near to each other on the circle, 
or to be a secant, as has been said above. 

217. Prov. 71.—Any inscribed angle is half of the angle at 
the centre which includes the same arc between its sides ; or an 
inscribed angle is measured by half its included are (198). 

There may be three cases: 1st, when one of the chords form- 
ing the angle is a diameter; 2d, when 
the chords le on opposite sides of the 
diameter passing through their point of 
meeting; 3d, when they lie on the same 
side of that diameter. 

In the first case, that of the angle 
DAB, we have (88) DCB = DAB + 
CDA. But the triangle ACD being 
isosceles, DAB=CODA. Hence, DCB 
2A byormDAB = DCB. 

In the second case, that of DAF, we have DAF = DAB + 
BAF; but DAB=4DCB, and BAF =4BCF, by what has 
been shown; hence, DAF = $DCF. 

In the third case, that of EAD, we have EAD = EAB — 
DAB; but EAB = 4ECB, and DAB = 4DCB, by what has 
been shown; hence, EAD = ZECD. 

218. Cor. 1.—Since, as has been said, a tangent may be 





60 ELEMENTS OF GEOMETRY. 


considered as a secant, the two points of intersection of which, 
with the circle, are indefinitely near to each other, this propo- 
sition is evidently true of the angle formed by a tangent and a 
chord; thus the angle DAG is measured by half the are DFA, 
or is equal to half the angle ACD taken on that side on which 
it is greater than two right angles. 

219. Cor, 2.—An angle inscribed in a semicircle is a right 
angle; an angle inscribed in any other arc is acute or obtuse, 
according as the are is greater or less than a semicircle. 

220. Cor. 3.—Angles including equal arcs, or inscribed in 
equal ares, are equal. 

221. Cor. 4.—The sum of the opposite angles of a quadri- 
lateral inscribed in a circle is equal to two right angles, since 
the sum of the arcs which they include is equal to the whole 
circle. | 

222. Prop. 72.—An angle formed by two secants meeting 
inside the circle is equal to half the sum of the included ares ; 
and an angle formed by two secants, a secant and a tangent, 
or two tangents, meeting outside the circle, is equal to half the 
difference of the included ares. 

Let AB and CD be two secants, intersecting at E. Draw 
AD. Theangle AEC = ADC + BAD (88). But ADC is 
measured by $AC, and BAD by $BD (217). Hence, AEC is 
measured by $(AC + BD). 





Let ABC and ADE be two secants. Draw CD. The angle 
CAE = CDE — ACD. But CDE is measured by $CE, and 
ACD by 4BD. Hence, CAE is measured by 4(CE — BD). 

Let AFH be a tangent. Draw CF to the point of contact. 


ELEMENTS OF GEOMETRY. 61 


The angle CAF = CFH — ACF. But CFI is measured by $CF, 
and ACE by BF. Hence, CAF is measured by 4(CF — BF), 

Let AG be another tangent. GAC is measured by 4(GC — 
GB) and CAF by (CF — BF). But GAF = GAC + CAF, 
Hence, GAF is measured by $(GCF — GBF). 

223. Cor.—lf the difference of the arcs included between 
two secauts, a secant and tangent, or two tangents, is zero, that 
is, if these arcs are equal, these lines will (115) be parallel, 
since they cannot meet, and make an angle with each other 
equal to zero. And conversely, if two secants, a secant and a 
tangent, or two tangents, are parallel, they will intercept two 
equal arcs on the circle. Jor if not, through one of the points 
of intersection or contact of one of them with the circle, draw 
a line that will, with the other secant or tangent, intercept 
equal arcs; this line will be parallel to the other secant or tan- 
‘gent; but this is (84) impossible. 

224. Schol. 1.—The last two cases of the proposition may be 
considered as sufficiently proved by the second, a tangent 
being, as shown before, only a particular case of a secant. 

225. Schol. 2.—Prop. 71 is a particular case of Prop. 72, in 
which the secants meet on the circle, and one of the included 
ares is equal to zero. It often may happen that a proposition 
serving to establish another is simply a particular case of that 
other, as, for instance, Prop. 40 is of Prop. 42. | 

226. Pror. 73.—lf two secants to a circle intersect each 
other, the product or rectangle of the distances from their 
point of meeting, to the points where one secant meets the circle, 
will be the same as the product or rectangle of the same dis- 
tances on the other secant. 

There are two cases, according as the secants meet inside or 
outside of the circle, as represented in the two figures. They 
can, however, be treated together. Let the two secants cut the 
circle in A and B, and C and D respectively, and each other 
in E. Join AD and BC. 

The triangles AED, CEB are similar, having the angles A 
and C equal, and also D and B. In the second figure the an- 
gles A and C are equal, because BAD and BCD, which are 
what they lack of two right angles, are (217) equal, each being 
measured by BD. 


62 ELEMENTS OF GEOMETRY. 


The equality in the other case is of course directly proved by 
the same proposition, the equal angles being measured by half 
the same arc, either AC or BD. 





Hence, the triangles being similar, we have a nan aa or 
AE. BE = CE. DE. 

227. Cor. 1.—If one of the secants becomes a tangent, we 
have, FE being this tangent, AE.BE = FE*; or FE = 
V AE. BE; or FE is a mean proportional between AE and 
BE. : 
228. Cor. 2.—If both secants become tangents, we have, GE 
being the other tangent, GE? = FE*?; or GE = FE; or the 
distance on two tangents which intersect from their points of 
contact to their point of meeting is the same. 

229. Cor. 3.—Two, and only two, tangents, EF and EG, can 
be drawn to a circle from a point, E, outside. For a tangent 
from E is a line which makes with a line EH, from E to the 
centre, and a radius to the point of contact, a right-angled tri- 
angle; and these triangles must (60) all be equal. Hence, the 
angles which the tangents make with EH must be equal to the 
angle opposite the radius in this triangle; but two, and only 
two, lines can be drawn from E, making such angles with EH, 
one on one side, one on the other. Therefore two, and only 
two, tangents from any point outside to a circle; the distance 
from the point to their points of contact is the same, and, by 
the equality of the triangles, the line drawn from the point to 
the centre bisects the angle which the tangents make with each 
other. It also makes equal angles with the radii drawn to the 
points of contact. . 

230. Cor. 4.—Hence, if a line be drawn bisecting any angle, 


ELEMENTS OF GEOMETRY. 63 


a circle having its centre on this line, and tangent to one of the 
lines forming the angle, will be also tangent to the other; for 
the other tangent from the vertex of the angle will make the 
same angle on the other side of the line to the centre that the 
one already determined does on the one side. 

231. Cor. 5.—Hence, if lines be drawn bisecting two angles 
of a triangle, a circle described about their point of meeting, 
which is tangent to one of the sides, will be tangent to all three ; 
and the line bisecting the third angle will pass through its 
centre. This circle is said to be ¢nscribed in the triangle. 
Hence, the lines bisecting the angles of a triangle all meet 
in the centre of the inscribed circle. 

232. Cor. 6.—The triangle will be divided into three tri- 
angles by these lines bisecting its angles; these triangles have 
the sides of the original triangle for bases, and all have the 
same altitude, namely, the radius of the inscribed circle. The 
sum of their areas is equivalent to that of the original triangle, 
and equivalent to the sum of their bases multiplied by half 
their common altitude; hence, the area of any triangle ts 
equal to its perimeter multiplied by half the radius of the 
msecribed curcle. | 

233. Def—A. Regular Polygon is one which is formed by 
dividing a circle into any number of aliquot parts, and con- 
necting the points of division by straight lines. The polygon 
is said to be znscribed in the circle by means of which it is 
formed, and the circle cérewmscribed about it. 

The sides of a regular polygon are all equal (197), being 
chords of equal arcs; and the angles are equal (220), being 
inscribed in equal arcs. 

The sides of a regular polygon being equal, they are (203) 
equally distant from the centre of the circle (which is also 
called the centre of the polygon). Hence, if another circle be 
drawn with a radius equal to the distance of the sides of the 
polygon from the centre, it will (207) be tangent to the sides of 
the polygon. This circle is said to be énscrzbed in the polygon, 
and the polygon circumscribed about it. 

234. Def:—The distance of the sides of the polygon from the 
centre, or, in other words, the radius of its inscribed circle, is 
called the apothem of the polygon. Thesame expression may 


64 ELEMENTS OF GEOMETRY. 


be used for the radius of the circle inscribed in any polygon, 
regular or not, in which a circle can be inscribed. 

235. Cor.—The regular polygon of four sides is manifestly a 
square; for its sides are equal, and its angles are right angles, 
since each is inscribed in a semicircle. 

236. Prov. 74.—The side of a square inscribed in a cirele 
is a mean proportional between the radius and the diameter of 
the circle. | 

Let ABCD be asquare inscribed in a circle having its centre 

at E. In the right-angled triangle ABC, 
B BE is (84) a perpendicular to the hypothe- 
nuse, since AB = BC. . Hence (148), 

: , aE = ap MH AB= VAEAO 
237. Cor.—The area of the inscribed 
square is twice that of the square of the 
D radius; for we have AB*= AE. AC; but 
AC = 2AE; hence, AB? = 2AE*, Hence 

also, AB = V2. AE. 

238. Cor—The side of a square circumscribed about a 
circle is evidently equal to the diameter of the circle. 

239. Prov. 75.—The side of a regular hexagon inscribed in 
a corcle as equal to the radius of the circle. 

Let AB be a side of a regular inscribed hexagon. The 
angle ACB is one-sixth of four right 
angles, and one-third of two right angles; 
hence, the angles CAB and CBA together 
are equal (87) to two-thirds of two right 
angles. But CAB and OBA are equal, 
the triangle ACB being isosceles. Hence, 

E each is equal to one-third of two right 
angles, or is equal to ACB. Hence (52), 
AB = CB or CA, since ACB = CAB or CBA. 

240. Prov. 76.—The side of an equilateral triangle wn- 
scribed in a circle is a mean proportional between the radius 
and three times the radius. 

Connecting the alternate vertices of the inscribed regular 
hexagon of the last proposition, we have an inscribed equilateral 


triangle BEF. Draw AE and CE. The figure ABCE is (111) a 


B 


ELEMENTS OF GEOMETRY. : 65 
parallelogram. Hence (178), AC? + BE? = AB? + BC? + CE? 


+ EA? = 4AB? = 4AC*; hence, BE? = BAC, or 57 = an 
241. Cor. Hence, also, BE = =' 3 AC. 
242. Prov. 77.—The side of a regular decagon inscribed in 
a curcle is a@ mean proportional between the radius and the 
radius diminished by the side of the decagon. 
Let AB be the side of a regular decagon inscribed in a circle, 
the centre of which is C. 
The angle ACB is equal to one-tenth of 
four right angles, and one-fifth of two right 
angles. Hence, the sum of the angles CAB 
and CBA is four-fifths of two right angles ; 
and since these angles are equal, the triangle 
ACB being isosceles, either of them is two- aX 
fifths of two right angles. Now draw a line 
from B to a point D on CA, so as to bisect 
the angle CBA. The angles CBD and DBA are each equal to 
one-fifth of two right angles, being half of CBA; and there- 
fore they are each equal to ACB. The triangle CDD is there- 
fore isosceles, by Prop. 20; and the triangle DBA is also 
isosceles, being equiangular, and hence similar to ACB. Hence, 
C= AByand AD= AG—AB=—CB— AB. But (147) 
ae hence Ae re which was to be 
me AG.’ OB SeRR) iB? 
proved. Or by the similarity of the triangles DBA and ACB, 
ee CB AB CB 
“een? OB — AB AB AB OB 


243. Cor. 1.—Clearing the equation ARAB TAR of 


fractions, we have AB? = CB? — CB. AB; or AB’ + CB. AB 
BR": one solving this quadratic equation, AB = — 30B 


+ 7$CB? = i.e . CB = about 618 . CB. 
944, Oor, 2.—By means of this proposition, a decagon may 


5 —1 
= Of the 
2 
radius, and applying it ten times asa chord. The square and 
hexagon may be inscribed in a similar way. By connecting 


the alter nate vertices of the hexagon and decagon, an equilat- 
5 


as before. 








be inscribed in a circle by taking a line equal to 


66 ELEMENTS OF GEOMETRY. 


eral triangle and a pentagon may be inscribed; though the 
former may also be inscribed by taking a line equal to V3 times 
(241) the radius and applying it as a chord. Of course the 
ratio of the side of the pentagon, or any other inscribed polygon, 
to the radius could also be ascertained. 

245. Def.—A line divided as AC has been divided, is said 
to be divided in extreme and mean ratio. 

246. Prop. 78.—The area of a regular polygon, or of any 
polygon in which a circle can be inscribed, is equal to its 
perimeter multiplied by half the apothem. 

For the polygon may be divided by drawn lines from the 
centre to its angles into as many triangles as it has sides. These 
triangles all have the apothem for their altitude; consequently 
the sum of their areas, which is the area of the polygon, is 
equivalent to the sum of their bases, or the perimeter of the 
polygon, multiplied by half the apothem; just as the area of a 
triangle has been shown, in Prop. 73, Cor. 6, to be equivalent 
to its perimeter multiplied by half the radius of the inscribed 
circle. 

247. Cor. 1.—The area of a circle is equal to its perimeter, 
or curcumference, as it is called, multiplied by half the radius. 
For the area of the circle is equal to that of a circumscribed 
polygon of an indefinitely great number of indefinitely small 
sides. | 

248. Cor. 2.—The circumference of a circle is less than the 
perimeter of any circumscribed polygon; for its area is less; 
and if we divide the areas by the radius of the circle, the 
quotients will be the circumference of the circle and the peri- 
meter of the polygon. 

249. Prop. 79.—Jf a circle be divided into any number of 
aliquot parts greater than two, and tangents be drawn at the 
points of division, these tangents will form a regular circum- 
scribed polygon of as many sides as there are parts into which 
the circle has been divided. 

Let A, B, D be three of such points of division on the circle. 
Draw a line in the plane of the circle perpendicular to the 
radii at these points; they will be (207) tangent to the circle. 

Any two of them, AG, BH, drawn from two consecutive 
points, as A and B, will meet (114) at some point, E; for, by 


ELEMENTS OF GEOMETRY. : 67 


hypothesis, the angle ACB is Jess than two right angles, there — 
being more than two equidistant points of division; hence the 
sum of BAH and ABG is less than 
two right angles, since it is what the 
sum of CAB and CBA lacks of two 
right angles, and is consequently equal 
(87) to ACB. Hence, the tangents 
will form a polygon, having as many 
sides as there are tangents, or parts 
into which the circle has been divided. If now we draw lines 
CE, CF from the centre to two consecutive vertices of this 
polygon, these lines will (229) bisect the angles ACB, DCB; 

hence, the angles ECB and FCB are equal, since the eles 
ACB eat DCB, of which ECB and FCB are halves, are Apelk 

being measured by equal arcs. Hence, the egies EBC and 
FBC are (51) equal; and CE = CF. The same may be proved 
of any two consecutive lines from the centre to points of meet- 
ing of the tangents; hence all these lines are equal, and a 
circle may be circumscribed about the polygon which they 
form. 

Now, since the angle ECB = FCB, as has been shown, it is 
half of ECF; but it is also half of ACB, as shown before. 
- Hence, ECF = ACB. And the same may be proved of: the 
angle formed by any two lines drawn from the centre to con- 
secutive vertices of the polygon ; namely, that.it is equal to the 
angle between two of the radii down to the equidistant points 
of division on the circle. Hence, all the angles formed by lines 
drawn from the centre to consecutive vertices of the poly- 
gon are equal. Hence, the arcs are equal, corresponding to 
them in the circle ey enrined about the polygon. Hence, 
the polygon corresponds to the definition of a regular polygon, 
and it is circumscribed about the circle ABD, since its Bits 
are tangent to it. 

250. Cor. 1.—A regular inscribed polygon of the same num- 
ber of sides may be formed, all the sides of which will be 
parallel to the corresponding ones of the circumscribed poly- 
gon, by connecting the points at which the lines to the vertices 
of the circumscribed polygon cut the circle; and conversely, 
given a regular inscribed polygon of any number of sides, a 





68 ELEMENTS OF GEOMETRY. 


regular circumscribed one of the same number of sides may be 
formed, all the sides of which will be parallel to those of the 
inscribed one, by drawing tangents at the centres of the arcs 
of which the sides of the inscribed ones are chords. 

251. Cor. 2.—Regular polygons of the same number of 
sides, like the inscribed and circumscribed polygons of this 
proposition, are semilar,; for they manifestly correspond to the 
definition of similar polygons given in the enunciation of 
Prop. 50. Their perimeters are consequently to each other as 
any other corresponding lines, as, for example, the radii or 
diameters of their circumscribed or inscribed circles; and 
their areas are (182) to each other as the squares of their peri- 
meters or of any corresponding lines. 

252. Prop. 80.—The perimeter of any regular polygon of 
an even number of sides greater than four, circumscribed about 
a circle, rs equal to twice the product of the perimeters of the 
regular circumscribed and inscribed polygons of half the num- 
ber of sides, divided by the sum of these perimeters ; and the 
perimeter of the regular inscribed polygon of this same even 
number of sides 18 a mean proportional between the perimeter 
of this circumscribed polygon and that of the regular inscribed 
polygon of half the number of sides. 

Let AB be the side of a regular circumscribed polygon of 
2n sides, 2 being any number greater 
than two, so that 2” will be an even 

* number greater than four; and let D 
be its middle point, and point of con- 
tact with the circle. Draw a line from 
C, making an angle with the radius CD 
equal to ACB, or twice DCB; this 
line will (114) meet DB produced in 

” some point, since ACB is less than a 


A D B E 


right angle, being of four right angles. Let this point of 


meeting be KE. ED will be half the side of the regular cir- 
cumscribed polygon of » sides. Let the line CE cut the circle 
in F, and draw DF. DF will be the side of the regular in- 
scribed polygon of 2n sides. Drop from F a perpendicular 
FG on CD. FG will be half the side of the regular inscribed 


ELEMENTS OF GEOMETRY. 69 


polygon of n sides. Denoting, then, the perimeters of the cir- 
cumscribed polygons of n sides and of 2 sides by c and ¢’ re- 
spectively, and those of the inscribed polygons of n sides 
and of 2n sides Fs 4 and 2’ respectively, we have 
a 

AB=— > ED= 5, DF=5., FG =—. 
ED and FG are ee being both perpendicular to CD; 
hence, by the similarity of the triangles CGI, CDE, we have 
ED CE CE 
EGaaer oy 


EB ED—DB_ ED 29ED 
Bat (147) or op Np abe = DD eT PAB 7 
q ED 2ED ; : 
since DB = 4AB. Hence, 7G = canes ie Putting in 


this expression the values above of ED, IG, and AB, we have 


{= ea 1; or o'¢ = 2cz, — e'2; or ce + e's = 2x: or, lastly, 





a C 

a7 2c . 

¢= -; which was to be proved. 
c-- 2 


For the second part, denote the middle point of DI’, where 
CB cuts it, by H. The right-angled triangles BHD and DGF 
are similar, having the angles BDH and DIG equal on account 


. Deel Ge 4 

of the parallelism of ED and FG. Hence, DB DE But 
DF FG ; 

tie aU; and DB —4AB. Hence, AB = pr Putting 


in, as in the first part of the demonstration, the values of 


AB, DF,and FG, given above, we have std aa ae or 2’ = Vc’A, - 
which was to be proved. Files 

258. Cor.—The circumference of the circle round and in 
which the polygons are circumscribed and inscribed, may be 
obtained to any degree of approximation by means of this 
proposition. Beginning with the perimeters of any two similar 
polygons, one circumscribed, the other inscribed, we can com- 
pute those of the Rifcmnece hed and inscribed polygons of 
double the number of sides, which will differ less from each 
other than those which we began, as is evident, since ¢’ is less 
¢, and 2’ greater than ¢ (42). And the circumference of the 


70 ELEMENTS OF GEOMETRY. 


circle will always be intermediate between the perimeters of 
the circumscribed and inscribed polygons, as shown (248). 

By repeating this process, evidently a great degree of 
accuracy may be obtained in the measurement of the cir- 
cumference, though it never can be precisely given in terms 
of the radius or artis with which it is incommensurable, 
as is shown in the higher mathematics. 

After proceeding some few steps in the approximation, a very 
accurate value of the circumference may be obtained in the 


following way. eNOSYs 








Reverting to our formulas, we find that ¢—7= . ae 
Cit 
. *9 . 12 rey 12 
cb — 4 a ea’ — 4 
= —- = ——.(¢ — 2), and that ¢’)— 2 = 7 — — = —___ 
e+ ions r+ 4 a C C 


q if oT 
=— —(C —v?). 
G ( ) 
Now when the number of sides in the polygon is great, and 


c, c’, 2, and @’ are very nearly equal to the circumference and to 
| . 
g a 
. will be about = $,and— about = 1. Hence, 
a C 





each other, 
C 


in this case we shall have, with great approximation, ¢’— 74 = 
4(¢ — 2), or o= $(¢ + 2); and 7 —t= & — 7, orv¥=F(C 4%). 

Now ¢ is the largest 2 the four, c’ the next, a’ the next, and 
2 the smallest; and it appears from what has just been shown 
that in the case we are treating of c’ is the arithmetical mean 
between ¢ and 2, and 2’ the mean between c’ and 2; hence, 
the four will be arranged as follows: 


A Biola &D 





w 

re] 
‘ 
9S 
i>} 


AB, AC, AD, and AE representing 2, 2’, c’, and ¢ respectively. 

The value if z' exceeds that of 2 by e’— 2’, which is equal 
to #(¢@— 2); hence, the next value of 2, when the number of 
sides is again doubled, will exceed this by 4(c’— 2’), or ms(¢ — 2). 
The next value will in the same way be greater by ¢ of this 
amount, or by y4-(¢ — 2); and so on; ence we may write at 
once, with considerable accuracy, | circumference = 7 + (£4 34 
+ gz t+ zbp ete.) (C—) = t+ ii (¢—%)= =4 + 3(¢—2%); or 
when the number of sides in the circumscribed and inscribed 


polygons is great, the circumference exceeds the perimeter of 


— ewe 


ELEMENTS OF GEOMETRY. of 


the inscribed polygon by about one-third of the difference be- 
tween this perimeter and that of the circumscribed polygon of 
the same number of sides. 

Beginning with the circumscribed and inseribed squares, the 
values of the sides of which are given in Prop. 71, and taking 
the diameter as a unit of measure, we have ne following 
values of the perimeters of the circumscribed and inscribed 
polygons, and of the approximate circumference, calculated on 
the rule just given: 


Number} Circumscribed Approximate Cir- 


Inscribed Polygon. 




















of Sides. Polygon. cumference. 

+ 4-0000000 2°8284271 3°2189514 

8 3°31387085 3°0614675 3°1455478 

16 3°1825979 3°1214452 3°1418294 

32 3°1517249 3°1365485 3°1416073 

. 64 3°1441154 3°1403312 3°1415936 
128 3°1422236 31412773 





3°1415927 


The value of the circumference of a circle in terms of its di- 
ameter, or the number of times the circumference contains the 
diameter, has been calculated to 200 places of decimals. Its 
value to fifteen places is 3°141592653589793. It is, of course, 
the same for all circles; for any two circles may be regarded as 
similar regular polygons of the same indefinitely great number 
of indefinitely small sides; and hence their perimeters, or cir- 
cumferences, as they are called, will be to each other as any 
corresponding lines, as, for instance, their diameters ; or denot- 
ae ie circumferences by ¢ and c’, and the diameters by d and 
, / 
a, — wae OL fk — ©, 
a ded 

This value of the circumference of a circle in terms of its 
diameter is denoted in mathematics by the Greek letter 7. 
Two approximate values of it are worthy of mention ; namely 
= and a The latter is easily remembered by writing its 
numerator after its denominator, thus, 11,33,55. Its value 
reduced to a decimal, is 3°14159292+. The other is by no 
means so accurate, but is more convenient for practical use. 


%2 ELEMENTS OF GEOMETRY. © 


254, Prop. 81.—The area of a circle is equal to m tumes the 
sguare of the radius. 

For a circle is a regular polygon of an indefinitely great 
number of sides, and its apothem is the radius; hence, since 
its perimeter is equal to 7 times its diameter, its area is (246) 
equal to 7 times its diameter multiplied by half its radius, or 
to m times the square of the radius. 

255. Cor. 1—For w times the square of the radius, we may 
write $7 times the square of its diameter; or the area of a cir- 


rate 
cumscribed square is — times the area of the circle. 
vis 


256. Cor. 2.—The areas of circles or equal parts of their 
areas, such for example as the areas of sectors having equal 
angles at the centre, are to each other as the squares of the 
radii or diameters. Such sectors are called s¢meur sectors. 

257. Cor. 3.—The areas of segments formed by connecting 
the ends of the radii of similar sectors by chords are equal 
to the areas of the sectors diminished by the areas of similar 
triangles; and the areas both of the sectors and of the triangles 
being as the squares of the radii, which are corresponding sides 
of the triangles, it follows that the areas of the segments are 
also to each other as the squares of the radii. Such segmeuts 
are called semzlar segments. ‘Their areas are of course equal 
parts of the areas of the circles; otherwise they would not be 
to each other as the areas of the circles are to each other. 


1S Oye elas Vel. 


MUTUAL RELATIONS OF LINES AND PLANES. 


258. The properties of planes were developed in the begin- 
ning as far as was necessary to ascertain those of figures formed 
in planes, and of parallel lines; we will now proceed to a 
farther investigation of them. 

259. Prop. 82.—/f an axis or perpendicular be dropped 
from a point outside on a plane: 

1. Lt ts shorter than any line that can be drawn from the 
pont to the plane. 

2. Lines drawn from the points outside to points on the 
plane, equally distant from the foot of the perpendicular, are 
equal. 

3. Of two lines drawn from the point outside to points on 
the plane, unequally distant from the foot of the perpendicular, 
the one drawn to the more distant point is the longer. 

1 comes from Prop. 13. 

2 is proved as 2 of Prop. 18, by means of Prop. 7. 

To prove 3, lay off a point on the line connecting the foot of 
the perpendicular with the more distant point, at a distance 
from the foot of the perpendicular, equal to that of the less 
distant point. The third line drawn from the point outside to 
the point thus determined will be equal to that drawn from the 
point outside to the less distant point, by 2 of this proposition ; 
but it will be less than that drawn to the more distant point, 
by 3 of Prop. 13; hence, the line drawn to the more distant 
point is greater than that drawn to the less distant point. 

260. Cor. 1.—All the points in a plane which are equally 
distant from a point outside, lie in a circle which has the foot 
of the perpendicular on the plane from the point outside for 
its centre; and they completely make up this circle. The 


74 ELEMENTS OF GEOMETRY. 


radius of the circle is (173) the square root of the difference of 
the squares of the distance of the points from the points out- 
side, and the perpendicular on the plane from that point. 

261. Cor. 2.—Hence there cannot be more than two points 
in any plane which are at the same definite distance from one 
point outside, and at the same or another definite distance from 
another point outside not on the same axis with the first point ; 
for (211) two circles which have not the same centre cannot 
intersect in more than two points. 

262. The perpendicular from a point on a plane is called the 
distance of the point from that plane. 

263. Prov. 838.—/f two lines be drawn in space, not lying m 
the same plane, a line connecting them can be drawn so as to 
be perpendicular to both. 

Let AB, CD, be two such lines. Drop from any point, E, of 
. CD, a perpendicular EF, on AB, and form 

a plane MN by rotating EF round AB. 
Drop on this plane a perpendicular, CG. 
This perpendicular will not coincide with 
CD, for it must be, by definition, parallel 
to AB, and hence must lie (70) in the 
same plane with AB; but CD, by the 
supposition, does not lie in the same plane 
with AB. 

Draw EG, and let fall on it (prolonged 
if necessary) a perpendicular, FH. Draw HK parallel to GQ, 
meeting CD (114) in K. K is not on AB, since AB and CD 
are not in the same plane, and hence, do not intersect. Drop 
KL perpendicular to AB. 

FH, being perpendicular to EG, and also to HK (since HK 
is an axis of MN), is an axis of the plane of CEG. But LK is 
parallel (71) to FH, since both are perpendicular to AB, and 
lie in the plane of the parallels AB and KH. 

Hence, LK is also an axis of the plane of CEG, and is con- 
sequently perpendicular to CD. But it has been drawn per- 
pendicular to AB; hence it is the line which it was to be shown 
could be drawn. 

264. Schol_—This proposition is the complement of Prop. 381. 

265. Prov. 84.—Through any two lines which do not wmter- 





ELEMENTS OF GEOMETRY. 15 


sect, two parallel planes can be passed, and only two parallel 
planes, unless the lines are parallel. 

For a line can be drawn (263) perpendicular to both of the 
lines, and connecting them; if two planes can be described 
about this line as an axis, at the points where it meets the other 
lines, these planes will be parallel, and will include those other 
lines. 

Only two parallel planes can be passed, unless the lines are 
parallel. For let the lines be AB and 
CD,.and suppose two parallel planes, as 
represented in the figure, to be passed 
through them. Draw axes to the planes 
from every point of the line AB; these 
will (as in Prop. 32) form a plane inter- 
secting the plane of the other line in EF. 
Now EF cannot be parallel to CD, since 
(71) it is parallel to AB, which is by supposition not parallel to 
CD; hence (115), it will meet CD. The axis to the planes at 
G, the point of meeting, will then be perpendicular to both 
lines; but, by Prop. 81, there can be only one such perpen- 
dicular; hence, these planes are the planes already proved to 
exist in the first part of this proposition, and there can be no 
other ones. 

_ 266. Prov. 85.—Two planes which are not parallel, will 
meet, uf sufficiently extended. 

~ Parallel planes were defined (104) as those which have a 
common set of axes; that is, any axis of one of two parallel 
planes is also an axis of the other. If 
two planes are not parallel, no axis of 
one will be an axis of the other, since if 
one were so, all would be. Drop then 
from any point, A, of one of the planes 
a perpendicular, AB, on the other. 

AB will be the axis at B of the plane 
in which B lies. Let AC be that part 
of the axis at A of the plane in which A les, which lies on 
the other side of it from AB. BAC will not form one straight 
line, since, as has been remarked, no axis of one of the planes 
is an axis of the other. 





Cc 





76 ELEMENTS OF GEOMETRY. 


Pass a plane (13) through the lines AC and AB. Draw in 
this plane lines AD and BE perpendicular to AC and BA at 
A and B respectively, on the side on which BAC is less than 
two right angles. ‘These lines will lie in the original planes in 
which A and B respectively lie. But these lines (114) will 
meet in some point, I’; since ABE is a right angle, while 
BAD = BAC — DAC, is less than a right angle. Hence the 
planes in which these lines lie will have at least this common 
point, F, in which they will meet. 

267. Prov. 86.—T wo planes which are not parallel will have 
more than one common point, and their common points will 
form a continuous and indefinitely long straight line. 

For let AB, as in the last figure, be a perpendicular from 
one of the planes on the other, and let 


4 AC be an axis of the plane in which A 
les; and let AF and BF be drawn also 
A as in the last proposition, in the plane 
BAC perpendicular to AC and BA re- 

B G spectively. 


Let GF be the axis at F of the plane 
of AFB. We shall have (173), AF? = 
AB? + BE*, and AG? = AF? + FG?: hence, AG* = eaiiee 
BF? + FG*. But BF? + FG? = BG’; hence, AG* = AB? + 
BG’; hence (175), ABG is a right angle, and G is in that one 
of the original planes in which B lies. 

Drop BH perpendicular to AF. BH will be parallel to AC 
(71), and will be therefore an axis of that one of the original 
planes in which A lies. Now we have (173), BF? = BH? + 
HF’, and BG? = BF’ + FG’; hence, BG? = BH? + HF? + 
FG’; but HE* + FG* = HG’, hence, BG? = BH*® 4+) iG 
hence (175), BHG is a right angle, and G is in that one of the 
original planes in which A lies. G, therefore, is in both of the 
original planes, and is a point of their intersection. But any 
other point of the line GH, produced indefinitely in either di- 
rection, could also be proved to be in both planes. Hence, the 
whole line GH. is in both planes. No point outside of the line 
GH can be in both planes; for, through any two points of the 
line GH, and a point outside that line, only one plane (65) 
can be passed. 


as 


ELEMENTS OF GEOMETRY. 1% 


268. Schol.—The sum of all the axes dropped from all the 
points of a line on any plane, or, in other words, the surface 
formed by the axis of the plane moving along this line, as in 
Props. 32 and 83, is continuous, by the principle of continuity, 
and has been shown in Prop. 32 to be a plane; and it was evi- 
dent in Prop. 88 that its intersection with the original plane 
was a straight line. The same was also evident in Prop. 382, 
by Prop. 25, Cor., and the principle of continuity. 

That the intersection of these two planes is a continuous 
straight line is a particular case of the proposition just proved. 

269. Prov. 87.—All pairs of lines, drawn like AF and BF 
in the last proposition, im two planes, and perpendicular to 
their intersection at the same point, make the same angle with 
each other. 

Let AF and BF, KG and LG be two such pairs of lines 
drawn perpendicular to FG in the 
planes, the intersection of which is C 
FG. Lay off any distances, as AF and 
EF, on the first pair, and lay off KG , 4 
equal to AF’, and LG to EF on the W\ 
other pair. Draw AK and EL. The an 
figures AFKG and EFGL are parallelo- L 
grams, by definition; hence, AK and 
EL are equal and parallel to GI’; and consequently they are 
equal and parallel to each other. 

Hence, AELK is a parallelogram by definition; hence, AE = 
KL. Hence, the triangles AI*E and KGL are (48) equal ; hence, 
the angles AFE and KGL are equal. 

This proposition, however, can be proved without any propo- 
sition regarding parallels. 

Let BAC and HGK be two angles ° 
made by two such pairs of lines; and EH 
let D be a point of the intersection AG 
of the planes, half way between A and 
G. D 

Let DE and DF be lines drawn at ‘3 
D, as AB and AC are drawn at A, or GH and GK at G. 

Reverse the line DA so that it falls on DG, taking with it 
the lines AB, AC, DE,and DF, and revolve these last round 


NN 


78 ELEMENTS OF GEOMETRY. 


the coinciding lines DA and DG till AB falls on GK; then 
will DE fall on the former position of DF, since it is in the 
plane of AB and AD; and DF will fall on the former position 
of DE. Hence, AC will fall on GH, since it is in the plane of 
DF and DA; hen the angle BAC coincides with, and is 
equal to, HGK. 

270. (on —These angles are epual (138) to those formed by 
the axes of the planes, “BA produced and AC, which are re- 
spectively perpendicular to BF and AF, and lie in the plane of 
BF and Al’, as shown in the last proposition. And as lines 
like AF and BF can be drawn, as in the last proposition, for 
any two axes meeting in a point, it follows that the angles 
made by any two such axes are equal for the same two planes. 

271. Def—This angle formed by the axes, or by lines drawn 
as in this proposition, is called the angle which the planes make 
with each other, or the diedral angle. Two planes, like two 
lines meeting, of course, make four diedral angles ; the opposite 
ones are Parle and the sum of adjacent ones patel to 2R, by 
Prop. 6. 

272. Def—When the diedral angle of two planes is a right 
angle, the planes are said to be perpendicular to each other. 

273. Prop. 88.—A plane including an axis of another plane 
as perpendicular to that other plane ; and, conversely, a plane 
which is perpendicular to another includes the axis of the other 
which is drawn from any of the points of the first plane. 

Let the plane AB include the axis CD of the plane EF; it 
will be perpendicular to EF. 

For draw CG in EF perpendicular to 
the intersection CB, then will DCG be 
the diedral angle, CD being also per- 
pendicular to CB as to any other form- 
ing line of EF. But CD is also perpen- 
dicular to CG, as it isto CB; hence, 
DOCG, the diedral angle, is a right angle, and the plane AB is 
perpendicular to EF. | 

Conversely, if AB is perpendicular to KF, it includes the 
axis of EF’, which is drawn at any point, as C, or D, of AB; 
for, draw CD in the plane AB, and CG in the plane EF, each 
perpendicular to CB. DCG is a right angle, since it is the 





ELEMENTS OF GEOMETRY. to 


diedral angle, and the plane AB is, by hypothesis, perpendicu- 
lar to EF. But DCB is also a right angle by construction ; 
hence, CD is the axis of EF at C. 

274. Cor. 1.—Each of the three lines CB, CD, and CG, is 
perpendicular to the other two, and consequently to their 
plane; and hence, by this proposition, any plane including 
either one of the lines is perpendicular to the plane of the 
other two. Hence, of the three planes, formed by combining 
these lines, two and two, any one is perpendicular to both of 
the others. This is always true for any three lines perpendicu- 
lar to each other. 

275. Cor. 2.—If another plane, perpendicular to EF, inter- 
sects AB, the intersection will be an axis of EF. For drop, 
from any point, H, of the intersection of this plane with AB, 
perpendiculars on its intersection with EF and on CB, the 
intersection of AB with EF. These perpendiculars must be 
axes of EF’ by the second part of the proposition. Hence, they 
must (72) coincide, since they pass throngh the same point H. 
But they lie in the new plane and in the plane AB respectively ; 
hence (267), they must be the intersection of these planes, 
Ilence, the intersection of these planes is an axis of EF. 

276. Def—The foot of a perpendicular dropped from a 
point on a line or a plane is called the projection of the point 
on the line or the plane, and the distance between the feet of 
two perpendiculars dropped from two points on a line or a 
plane is called the projection of the distance between the two 
points on the line or plane. 

Similarly the line formed by the perpendiculars dropped 
from any line without defined extremities on a line or a plane is 
called the projection of the line on the line or plane. If a 
line is projected on another line, the two lines need not be in 
the same plane. 

277. Cor. 4.—By what has been shown, it is plain that the 
projection of a line with or without defined extremities on a 
plane, lies in the same plane with the line, and that this 
plane is perpendicular to the one on which the projection 1s 
made. 

978. Cor. 5.—It is also plain (173) that the square of any 
line is equal to the sum of the squares of its projection on any 


80 ELEMENTS OF GEOMETRY. 


plane, and of the difference of the distances of its extremities 
from that plane, the distance of one of these extremities being 
tuken as negative if they are on opposite sides of the plane. 

279. Prop. 89.—ILf a plane intersect two parallel planes, its 
intersections with them are parallel to each other. 

lor these intersections both lie in the intersecting plane; 
hence (115), if-they are not parallel they must meet in some 
point ; but if they meet in any point, the parallel planes in 
which they lie inust meet in that point; but this is impossible, 
since (103) parallel planes are everywhere equally distant from 
each other. 

280. Prop. 90.—Parallel lines connecting parallel planes 
are alt equat to each other. 

For, pass a plane through any two of the parallel lines, the 
intersections of this plane with the parallel planes are (279) 
parallel; and hence, with the parallel lines, they form (111) a 
parallelogram ; hence, the parallel lines are (106) equal. 

This proposition may be stated as follows: the distance be- 
tween parallel planes measured on any parallel lines is the same, 
whether these parallels are perpendicular to the planes or not. 

281. Def—A line is said to be parallel to a plane when it 
lies in a plane which is parallel to that plane. 

It follows from this definition that a line cannot meet a plane 
to which it is parallel. 

It follows also that a line which is ea to any plane is 
parallel to all the set of parallel planes to which that plané 
belongs; for a plane can, by definition, be passed through it 
which will be parallel to that plane, and therefore to the whole 
set; and, therefore, the line lying in it will be parallel to the 
whole set. 

282. Prop. 91.—A line which is not parallel to a plane will 
meet that plane if sufficiently produced ; and it will meet at 
en only one point. 

Let AB be the line and CD the plane. Project AB on the 
plane CD, and let FG be the projection, and F the projection 
of some point, E, of the line AB. 

Now EF is not perpendicular to AB; for if it were, AB 
would le in a plane parallel to.the original one; but it is per- 
pendicular to FG; EB will consequently (114) meet FG in 


ELEMENTS OF GEOMETRY. 81 


some point; and at this point it will, therefore, meet the plane 
CD, in which FG lies. Jt will (49) meet it in only one point. 

283. Cor. 1—A line, HK, which is 
parallel to a line, FG, of a plane, CD, is 
parallel to that plane. (IG is not sup- 
posed to be the projection of HK in 
this and the following corollary.) For 
if not, it will meet that plane, by this 
proposition; but if it meets it, draw 
through the point of meeting a linein é¢ 
the plane parallel to FG. This line 
being parallel to FG, must be parallel to HK ; but by hypothe- 
sis it meets it; the hypothesis therefore that HK is not parallel 
to the plane is absurd. | 

284. Cor. 2.—A line, HK, parallel to a line, FG, of a plane, 
CD, is parallel to the intersection with CD of any plane in 
which HK les. For it les in the same plane with that inter- 
section, and if not parallel to it, must (114) meet it; but if it 
meets it, it meets the plane CD. But this cannot be, by Cor. 1, 
for it is parallel to CD. 

285. Defi—The acute angle made by a line with its pro- 
jection on any plane is called the angle of the line with the 
plane. The sum of this angle and the acute angle made by the 
line with the axis raised at the point where the line meets the 
plane is evidently equal to one right angle. 

286. Prop. 92.—If two angles, lying in different planes, 
have their sides parallel and lying on the same side of the line 
joining their vertices, these angles 
will be equal, and their planes will ™ 
be parallel. | 

Let BAC and EDF be two such 
angles lying in the planes MN and 
PQ; AB and DE being on the same 
side of AD, and this also being the 
case with AC and DF. 

Drop from A a perpendicular, AG, 
on the plane PQ. Draw in this plane 
GH parallel to DE and GK to DF. 
GH will be parallel to AB and GK to AC. But GH. and GK 

6 








82 ELEMENTS OF GEOMETRY. 


are perpendicular to AG, since AG is an axis of the plane 
PQ; hence, AB and AC are also perpendicular toAG. Hence, 
the angles BAC and HGK are equal, each measuring the angle 
of the planes AH and AK. But HGK is equal (138) to EDF; 
hence, BAC and EDF are equal. And the planes MN and PQ 
are parallel, since AG is an axis of MN as well as of PQ; AB’ 
and AC being perpendicular to it, as has just been shown. 

25.( prs it —If at every point of either of two lines which 
do not meet at all, lines be drawn parallel to the other one, and 
then if at every hain of the other one, lines be drawn parallel 
to the first one, by this proposition every one of the angles 
formed on the second line will be equal to every one of the 
angles formed on the first, and the constant angle thus formed 
is said to be the angle which the two original lines make with 
each other. Or, what is meant by the angle formed by two 
lines which do not meet is the angle between either of them 
and a line drawn at any point of it parallel to the other. Thus, 
for instance, in Prop. 86, FG is said to be perpendicular to AC 
or to BA, or to any axis of any plane in which IG lies ; because 
an axis drawn to such a plane at any point of IG would be 
parallel to any other axis, and would be perpendicular to FG. 
The intersection of two planes is then perpendicular to all 
their axes. 

288. Cor. 2.—By Prop. 89, and this proposition, we con- 
clude, that if two intersecting planes, BAD and CAD, cut two 
parallel planes, MN and PQ, the sides of the angle formed by 
their intersections in one of these planes will be parallel to the 
sides of that formed in the other, and that the angles will be 
equal. 

289. Cor. 3.—Two planes, connected by three equal parallel 
lines, are parallel. (By Prop. 48, and this proposition.) 

290. Prop. 938.—If any number of lines arecut by any 
number of parallel planes, they will be dinided proportionally ; 
that is, the part of any one of the lines between any two of the 
planes will be the same multiple of the part included between 
any other two, that the part of any other line included between 
the same two planes is of the ge included between the same 
other two. 

Let AB, CD be any two of the lines, and EF, GH, any two 


ELEMENTS OF GEOMETRY. 83 


of the planes, the extreme ones, for instance, of the series of 
four which may be considered. | 

Draw AD. The intersections of the plane ABD with the 
four planes are (279) parallel; those of 
the plane of ACD are parallel, by the 4 
same proposition. Hence (124), the ie PS 
part of AB or of CD between any two 
of the planes is the same multiple of 
the part between any other two that the 
part of AD between the same two of 
the planes is of the part between the Sar 
saine other two; and hence, the part H 
of AB between the first two planes is 
the same multiple of the part between the other two that the 
part of CD between the first two is of the part between the 
other two. 

291. Prop. 94.—A triangle can always be moved from any 
one position in space, to any other position, by two movements, 
one of translation, the other of rotation. 

First move the triangle by a movement of translation (137) 
such that one of its vertices shall fall on the new position in- 
tended for it. Then connect the present and intended po- 
sitions of the other two vertices by straight lines; and pass 
through these lines two parallel planes (265). If the two 
lines intersect, the two planes will coincide. Project these two 
lines connecting the present and intended positions, on that 
one of the set of parallel planes to which these two planes 
belong, which passes through the point C, which is already in 
its intended position. On account of the parallelism of the 
planes, the lines connecting the present and intended positions 
of the other two points will be projected in their true length; 
and the projections of the lines connecting C with the present 
and the intended positions of either of the other two points will 
be (278) equal, since these present and intended positions are 
equally removed from the plane of projection. 

The projection of the line connecting the present positions 
of the other two points will also be equal (278) to the projection 
of the lines connecting their intended positions, since the 
difference of distance of the extremities of these lines from 


84 ELEMENTS. OF GEOMETRY. - 


the plane of projection is equal to the distance of the two: 


parallel planes first passed from each other. 


Let A and B, then, be the projections of the present Eee 


of the other two points, and A’ and 
B’ those of their intended positions ; 
we have AA’ and BB’ equal to the 
actual distances of the present po- 
sitions from the intended ones; and 


=- A’B’. The triangles CAB and 
CA/B’ are then (48) equal, and the 
ee ACB isequal to the angle A’CB’. 





Fig. 1, Fig. 2. - Now there may be two cases; either. 


these angles lie on the. same side of CA and CA’ renee 
or on opposite sides. In the first. case, represented by Fi fig. 1, 
the angle ACA’ is equal to the angle BOB, and a rotation oe 
the triangle around the axis of the parallel planes passing 


we have CA = CA’, CB = CB’, AB 


through C through an angle equal to ACA’ or BCB’ will bring. 
the projections of its vertices, and consequently the vertices. 


themselves, into their intended positions. 
In the second case, represented by Fig. 2, lines drawn to the 


middle points of AA’ and BB’ will (82) bisect the angles 


ACA’ and BCB’, and will consequently coincide; and hence, 
since these coinciding lines are (82) perpendicular to AA’ and 
BB’, it follows that AA’ and BB’ are (71) parallel. And since 
AA’ and BB’ are (279) parallel to the lines of which they are 
the projections, these lines, that is to say, the lines connecting 


the present and intended -positions of the other two points of 


the triangle, must (85) be parallel, and. hence must lie in the 


same plane. 


Let A, A’, B, B’, then, in Fig. 2, now represent the unpro- . 
jected pr ister ae sikeatiled positions of the other two vertices, | 
and C the vertex which has arrived in position, and otha 
perhaps, does not lie in the plane of AA’ and BB.’ Unless AB’ 


and A’B’ are parallel, they will (114) meet in some point, D, of 
this plane, and the triangle ACB can be put in the position 


“= 


A’CB’ by rotating it around AD. If they are parallel, the. 


rotation must be.around a line drawn through C parallel to them. 


292. Cor, 1.—Any system of points may be put in any new 


ELEMENTS OF GEOMETRY. 85 


position in space by one translation and one rotation, since when 
three of its points are in their new positions, the remainder 
also must be. 

293. Cor. 2.—It is of course possible to move a triangle or a 
system from one position to another by an infinite variety of 
translations, provided the corresponding rotation is given to the 
triangle or system afterward. For it is plain that not only can 
any one of the points conceived of as forming the system (as, 
for instance, the three vertices of the triangle) be moved to its 
new position, and the rest of the points translated with it, but 
that any other point whatever, bearing any position with regard 
to the points properly belonging to the system, can be conceived 
as admitted into the system, for the time, and moved to a point 
bearing the same position with regard to the new positions of 
the other points, the rest of the system being translated with it, 
as in the previous case ; then, when this point, whether properly 
belonging to the system or not, has reached its position, a ro- 
_ tation can be made so as to bring the rest into position. 

294. Cor. 3.—Any motion whatever which can be given to 
a system of points, can evidently be considered as composed of 
a translation of any one of them through its successive positions, 
and a rotation around an axis passing through this point; the 
direction of the axis may change, and the translation may be 
ona curve ; that is, it may be composed of an infinite number of 
‘infinitely short straight translations, for every one of which 
there is an axis of rotation, perhaps remaining fixed in direction 
or perhaps not remaining fixed, but changing either occasionally 
or continually ; in which cases, respectively, it may be called the 
permanent, temporary, or instantaneous axis of rotation. 

295. Cor. 4.—As the motion can be reversed, it is evident 
that the rotation may, if we please, precede the translation. 


BOO Wits 


STRUCTURES IN SPACE. 


296. Def—If a plane polygon be translated along straight 
lines, from one position to another, lying in a different plane, 
and planes be passed through the lines of translation of the 
adjacent vertices, the resulting structure of planes is called a 
prism. 

The planes passing through the lines of translation of the 
adjacent vertices are parallelograms, as shown in Prop. 48, and 
will include all the intermediate positions (70) of the sides of 
the polygon in the translation, since these intermediate po- 
sitions are all parallel to their initial positions. These planes 
are called the s¢des of the prism; their sum is called its convex 
area. The initial and final positions of the polygon are called 
the bases of the prism. The bases lie (289) in parallel planes. 
The distance between them is called the al¢etude of the prism. 
The equal lines of translation of the vertices of the polygon, or 
the intersections of the sides of the prism, are called its edges. 

If the polygon is translated along the axes of its plane, that 
is, if the edges are perpendicular to the bases of the prism, it is 
called a reght prism; if not, it is called an oblique prism. The 
sides of a right prism are evidently rectangles, since the axes of 
a plane are perpendicular to lines passing through them in the 
planes. If the base of a right prism is a regular polygon, it is 
called a regular prism. 

The perimeter of a prism is a line passed through all its 
faces, in the plane perpendicular to the edges. It is the peri- 
meter of the polygon formed by the intersection of this plane 
with the faces of the prism. It is also the sum of the altitudes 
of the sides of the prism, if the edges are taken as their bases. 

297. Def—If the number of sides of a regular prism is in- 





ELEMENTS OF GEOMETRY. 87 


creased indefinitely, keeping the circle inscribed in its base un- 
changed, it is called a cylinder. The base coincides with the 
inscribed circle ; and the cylinder may be conceived as formed 
by the rotation of a rectangle having a base equal to the radius 
of this circle, and an altitude equal to the altitude of the 
cylinder, around the side which is equal to this altitude. 

The perimeter of a cylinder is more properly called its cir- 
cumference. 

298. Prop. 95.—The convex area of a prism rs equal to the 
rectangle of rts perimeter and rts edge. 

For this area is composed of those of parallelograms, each 
one of which is equal to the rectangle of the base and altitude 
of the parallelogram ; but (taking the edges as bases) their bases 
are all equal, and the perimeter of the prism is the sum of their 
altitudes ; hence, our proposition is proved. 

299. Cor.—The convex area of a right prism, or of a cylinder, 
is equal to the rectangle of the perimeter of either base and its 
altitude. | 

300. Prov. 96.—Jf the sides of a prism are cut by parallel 
planes, the intersections form equal polygons. 

For the portions Ab’, AC’, etc., of the faces of the prism in- 
cluded between the parallel planes are paral- 
lelograms (111), since the edges AA’, BB’, ete., 
are parallel, and (279) A’B’ is parallel to AB, 
A’C’ to AC, ete.; hence, A’B’ is equal to AB, 
A’C’ to AC, ete., and (286) the angle B’A’C’ 
is equal to the angle BAC. 

In the same way all the other sides and 
angles of either polygon can be shown to be 
equal to the sides and angles of the other 
polygon respectively, taken in the same order; hence the poly- 
gons are equal. 

301. Schol.—A prism may be conceived as indefinite, having, 
that is to say, definite sides, but no bases; in that case, a 
definite prism can be formed by cutting the indefinite one by 
any two parallel planes, which will form two equal bases, as 
has been shown; or by arresting the indefinite translation by 
which the indefinite prism is formed, at two points. 

302. Def—A. Parallelopipedon is a prism the bases of which 





88 ELEMENTS OF GEOMETRY. 


are parallelograms. If the bases of a right parallelopipedon 
are rectangles, it is called a rectangular parallelopipedon. The 
rectangular parallelopipedon the base of which is the square 
of its altitude, is called a cube. 

303. Prop. 97.—TLhe opposite sides of a parallelopypedon 
are parallel and equate. 

Let AH, BG, be two opposite sides of a parallelopipedon, 

the bases of which are AC and EG. 
as We have AD, DH, HE, and EA paral- 
Ee * lel and equal to BC, CG, GF, and FB 
respectively, since the bases and sides are 
all parallelograms ; hence, on account of 
the parallelism, the angles of the paral- 
lelogram ATH are (286) equal to those of 
et ee x BG, and their planes are parallel, and 
on account of the equality, the sides of 
the parallelogram AH are equal to those of BG; hence, the 
parallelogram AT, having its angles and sides all equal to 
those of BG, taken in the same order, is equal to it in every 

respect. 

304. Cor. 1.—Any two sides of a parallelopipedon may be 
taken as its bases. 

805. Cor. 2.—The square of any diagonal of a rectangular 
parallelopipedon is equal to the sum of the squares of the three 
edges corresponding to its three different pairs of bases. 

Let AG be a diagonal of a rectangular parallelopipedon; we 

have (173) AG? = AO? + CG’, and AC? = 
ae AB? +. BC, Henee, AG* = Alb? BG 

=a E CG’; or, as we may write it, AG? = AB? + 

AD? + AE? 
This corollary may be stated as follows: 
The sum of the squares of the projections of 
|,» any line on three lines at right angles to each 
——s - other, is equal to the square of the line atself. 
It is immaterial whether the line intersect 
the lines on which it is projected or not, and also whether the 
lines at right angles to each other intersect each other or not. 
(See Prop. 92, Cor. 1.) For we can draw at one extremity, A, 
of the line AG to be projected, three lines, AB, AD, AE, 


ELEMENTS OF GEOMETRY. 89 


parallel to the three at right angles to each other; these three 
lines so drawn will be (286) at right angles to each other, and 
we shall have AG? = AB? + AD? + AE’; but the projections 
of any line on any parallel lines are equal, for they are the 
distances between parallel planes passed through the extremi- 
ties of the line, and having the parallels for their axes; hence 
for AB, AD, and AE, we may put the projections of AG on 
the original lines respectively. ) 

306. Cor. 3.—We may add that the projections of a line on 
parallel planes are also equal; for these projections are (279) 
parallel. j 

307. Def—lf all the points of a plane polygon be moved 
along straight lines, meeting in a common point outside of its 
plane, and planes be passed through the lines of movement of 
the adjacent vertices, the resulting structure of planes is called 
a pyramid. 

The planes passing through the lines of movement of the 
adjacent vertices are triangular, and are called the sedes of the 
pyramid; their sum is called its convex area. The initial po- 
sition of the polygon is called its base. The common point to 
which the polygon moves is called the vertex of the pyramid. 
The perpendicular from the vertex on the plane of the base is 
called its altaztude. The lines on which the vertices of the 
polygon move, and which meet in the vertex of the pyramid, 
are called its edges. If the polygon is one in which a circle 
can be inscribed, and the altitude falls on the centre of this 
circle, the pyramid is called a rzght pyramid. If the base of a 
right pyramid is a regular polygon, it is called a regular 
pyramid. 

The perpendiculars from the vertex on the sides of the base, 
prolonged if necessary, are called the slant heights of the 
pyramid. In a right pyramid they are all equal, being tlie 
hypothenuses of equal right triangles, the other sides of which 
are the altitude of the pyramid and the apothem of its base. 
For the perpendiculars on the sides of the base from its centre, 
or the apothems of the base, fall (73) at the same points with 
those from the vertex of the pyramid. 

308. Defi—If the number of sides of a regular pyramid is 
increased indefinitely, the circle inscribed in its base remaining 


90 ELEMENTS OF GEOMETRY. 


unchanged, it is called a cone. The base coincides with the in- 
scribed circle; and the cone may be conceived as formed by 
the rotation of a right triangle, having one side about the right 
angle equal to the radius of this circle, and the other equal to 
the altitude of the cone, around this latter side which is equal 
to the altitude. 

The slant height and edge of a cone are the same. 

309. Prop. 98.—The convex area of a right pyramid is 
equal to half the rectangle of its slant height and the pervm- 
eter of its base. 

For the convex area is composed of those of triangles VAB, 
VBC, ete., having an altitude equal to the 
slant height of the pyramid, and _ bases 
equal to the sides of its base; hence, it is 
equal to the sum of the areas of these tri- 
angles, or half the rectangle of their com- 
mon altitude, which is the slant height of 
the pyramid, and the sum of their bases, 
which is the perimeter of the base of the 
pyramid. 

310. Cor.—The same is true for a cone, 
which is only a particular kind of right pyramid. 

311. Prop. 99.—/f a pyramid be cut by a plane parallel to 
ats base : | 

1. The edges and the altitude will be divided proportionally ; 

2. The section will be a polygon similar to the base. 

Let ABCDE — V be a pyramid, and VP its altitude; and 
let it be cut by a plane parallel to its base 
in A’B’C'D'E’, the altitude being cut in P’, 
VAS CB pV 
VA’ Vpr '® = ype 

For the triangles VAB and VA‘B’ are (279) 
similar; the same is trueof VBC and VBC, 
etc., and also of VAP and VA'P’, VBP and 
V B' Ps, ete; 

And the polygons ABCDE and A’B’C'D’E’ 
will be similar; for they are (286) mutually 


Ae? : AB | a3Vi 
equiangular ; and their sides are‘proportional, since NB = VB 





then will ete. 





ELEMENTS OF GEOMETRY. 91 


and ee also = ae Hence, fas = mae The same can 
be shown for the remaining sides. 

312. Cor. 1.—The perimeter of the section is to the perim- 
eter of the base in the same ratio as the sides, which ratio is 
that of the distances of the section and the base respectively 
from the vertex, the distance being measured (290) on any line 
passing through the vertex; and the areas of the section and 
the base are in a ratio which is the square of this. 

313. Cor. 2.—Hence, the polygon which forms the pyramid 
may be conceived as moving, so as to keep its plane always 
parallel to its initial position, and reducing its sides continu- 
ally in the ratio of the distance of its plane from the point 
to which it is tending ; or as starting from that point and 
developing its sides in the ratio of the distance of its plane 
from that point, and its area, consequently (181), in the ratio of 
the square of that distance. 

314. Def—A_ frustum of a pyramid is the portion of it in- 
cluded between its base and a section parallel to the base. 

It may be conceived as formed by an incomplete motion of 
the kind described in the last corollary, of the polygon toward 
the vertex. 

The section is called the wpper base of the frustum; the 
base of the pyramid, its lower base ; the distance between the 
planes of its upper and lower bases, its altitude; the distances 
between the corresponding sides of its upper and lower bases, 
its slant heights. 

A circle can (311) be inscribed also in the upper base of a 
frustum of a right pyramid; and hence the slant heights of 
such a frustum are all equal, being the difference of the slant 
heights of two right pyramids. 

Similarly a frustum of a cone is the portion of it included 
between its base and a section parallel to the base. It may of 
course be conceived as formed by the rotation of a trapezoid 
having two right angles about the side included between those 
angles. This side is the altitude of the frustum, the other side 
its slant height. 

315. Prop. 100.—The convex area of the frustum of a 
right pyramid is equal to the rectangle of its slant height 


92 ‘ELEMENTS OF GEOMETRY. 


and half the sum of the perimeters of its upper and lower 
bases. 

For this convex area is composed of the areas of trapezoids, 
each one of which is equal to the rectangle of the altitude of the 
trapezoid and half the sum of its upper and lower bases. But 
these altitudes are the slant height of the frustum; hence, the 
convex area of the frustum is equal to the rectangle of that 
slant height and half the sum of the upper and lower bases of 
the tr eas which last is half the sum of the per imeters of 
the upper and lower bases of the frustum. 

316. Cor. 1.—The same is true of the frustum of a cone, a 
cone being only a particular kind of right pyramid. 

317. Cor. 2.—The frustum being formed by the rotation of 
a trapezoid, it is clear, by Prop. 81, that the circle formed by 
the middle point of its slant height is equal to half the sum of 
the circumferences of its upper and lower bases; since this 
circle has for its radius what has been called (165) the mean 
breadth of the trapezoid, which is equal to the half sum of its 
upper and lower bases, the extremities of which form, by 
rotation, the bases of the frustum. 

Hence, the convex area of the frustum of a cone is equal to 
the rectangle of its slant height and its circumference half way 
between the bases; or, if a line be revolved about another in 
the same plane with it, the area formed is equal to the rectan- 
gle of the line and the circle formed by its middle point, if 
the revolving line does not cross the line around which it re- 
volves (247, 310). 

318. Def—A Sphere is a surface formed by the rotation of 
a semicircle around the diameter connecting its extremities. 

All the points of the semicircle being equally distant 
from the centre of this diameter, all the points of the sphere 
will also be so. ‘This centre is called the centre of the 
sphere. The radius and diameter of the semicircle which 
forms the sphere are called the radius and diameter of the 
sphere. | 

319. Prop. 101.—LHvery section of a sphere, made by a 
plane, is a circle. 

For let C be the centre of the sphere, and CA a perpendicu- 
lar let fall from C on the plane which makes the section, and 


ELEMENTS OF GEOMETRY. 93 


let B be a point of the section; CAB will be a right-angled 
triangle. 

But CB is the same for all points of the 
sphere; and CA is the same for all points 
of the section; hence (60), AB is the same 
for all points of the section, or, the section 
is a circle, having A for its centre. 3 

320, Cor, 1.—The radius of this circle is SE SY 
equal (173) to the square root of the differ- : 
ence of the squares of the radius of the sphere and the distance 
of the plane of the circle from the centre of the sphere; if, 
then, the plane of the circle passes through the centre of the 
sphere, the radius of the circle has its greatest pe value, 
and is equal to the radius of the sphere. 

321. Def—Such a circle is called a great circle of the 

sphere ; half of it is what, by rotation, forms the sphere. Any 
other section is called a Cau circle 4 the sphere. The great 
circle, being the longest line which can pass around the sphere 
in one plane, is said to measure the cirewmference of the 
sphere. 
_ 822, Cor. 2. Saye any two points on the anrface of the 
sphere, a great circle can be made. to pass, by passing a plane 
through the two radii of the sphere drawn to these points. 
Only one great circle can (65) be made to pass through these 
two points, unless they be at opposite extremities of.a diameter, 
in which case, any number of great circles can be passed 
through them. 

323. Cor, 3.—Through any three points on the surface of the 
sphere a circle, great or small, can be passed, and (65) only one. 

324. Def—A great circle divides the sphere into two por- 
tions which are manifestly equal, since the radius of the sphere 
being everywhere the same, they can be made to coincide. 
Sasa por tions are called Pees | 

A circle in general divides the sphere into two portions, 
which are Polece segments. 

The portion. of the sphere included between two circles, the 
planes of which are parallel, is called a zone. A segment is a 
species of zone, in which one of the circles reduces to zero; a 
hemisphere is a species of segment. | 


94 ELEMENTS OF GEOMETRY. 


The distance between the parallel planes of a zone or seg- 
ment is called its altetude. 

Propositions may be proved for the sphere corresponding to 
Props. 69 and 70 for the circle. For “a line” we must write 
“a plane,” and for “two points,” “a circle.” 

Definitions, similar to those for the circle (215) can also be 
given for the sphere. 

325. Prop. 102.— The area of a sphere ts equal to the rec- 
tangle of its circumference and diameter. 

Circumscribe half of some regular polygon having an even 
number of sides, around the semicircle, which, 
by rotation, forms the sphere, and rotate this 
with the semicircle. 

The areas formed by any one of the sides 
of this polygon will (817) be equal to the rec- 

K tangle of itself, and the circle formed by the 
point where it touches the semicircle inscribed 
in it; or (254) it will be equal to 27 times the 
rectangle of the side itself, and the distance of 

F the point where it touches the inscribed semi- 
circle from the axis of rotation. Or, in the 

figure, the area formed by the rotation of AB around EF 
is 2r7AB.CD,C being the centre of AB, where it touches 
the inscribed semicircle, and CD being the perpendicular from 
C on the axis. 

But if we drop perpendiculars AG and BH on EF, and a 
perpendicular AL on BH, the triangles BAL and CKD are 
similar (139); and also we have AL = GH; hence, 

aE = Gp AB. OD = CK. AL; 
and the area formed by AB is equal to 27CK . AL. 

The area, therefore, formed by the whole semi-polygon 
would be 27CK . EF, or (252), the rectangle of the inscribed 
circle and its diameter. 

But this result is independent of the number of sides of the 
polygon; hence, it is true when their number is increased to 
infinity ; but in this case, the semi-polygon coincides with the 
inscribed semicircle, and the area formed is that of a sphere, 
of which the inscribed semicircle is the semi-circumference ; 


n™ 


ELEMENTS OF GEOMETRY. 95 


and hence the area of the sphere is equal to the rectangle 
of its circumference and diameter. 

326. Cor. 1.—The circumference of the sphere being (253) 
equal to 2c times the radius, or mw times the diameter, the 
area is equal to 47 times the square of the radius, or 7 times 
the square of the diameter; or, since 7 times the square of the 
radius is (254) equal. to the area of a great circle, the area of 
the sphere is equal to the sum of the areas of four great circles. 

327. Cor. 2.—The area of a zone or segment is equal to the 
rectangle of the circumference of the sphere and the altitude 
of the zone or segment. 

328. Cor. 3—The areas of zones or segments on the same 
sphere, or equal spheres, are to each other as their altitudes. 


BO:O 7K SV Teh. 


VOLUMES IN SPACE. 


329. Def—A volume is the portion of space enclosed in a 
structure of planes or curved surfaces. It is called the volume 
of the structure which encloses it; thus the space enclosed 
within the planes of a prism is called the volume of the prism. 

A volume may always be conceived as formed by the move- 
ment of a plane area, the plane of which keeps always parallel 
to itself; though the figure limiting the area may undergo 
changes of form, and even to that extent that it may break up, 
or being broken up, may completely or partly unite. 

It is evident that two volumes are equal when one may be 
made to coincide with the other by a different arrangement of 
its parts, and that the sums or differences of equal volumes are 
equal. 

330. Prop. 103.—Two rectangles, which are equal in area, 
may be superposed by a different arrangement of the parts of 
erther area. 

Let ABCD, CEFG, be two rectangles equal in area, and 
placed so that they have a common angle C, 
and that the angles BCD and ECG are opposite 
to each other. 

Draw AF’, cutting CB and CE in M and N, 
and draw HL through C parallel to AF; and 
complete the rectangle DCGK. We shall then 
have (124), = == a or KL. KA =n Kes 
or, KL (AD + CG) = KH (EC + CD). 

But, since the rectangles ABCD and CEFG 
are equal in area, we have (157),CD.AD = CG. EQ; or, 
adding CD .CG to both members, CD (AD + CG) = CG 





-ELEMENTS OF GEOMETRY. oT 


(EC + CD); hence, = = es But by the similarity of the 
triangles LHK and LCG, oe = a hence, CD = LG. 

The triangles CHD and LCG are hence (60) equal; but the 
triangles AMB and NIE are (61) or (51) equal to CHD and 
LCG respectively ; hence, they are equal to each other, and 
capable of superposition, as are also CHD and LOG. And the 
areas of the parallelograms AMCH and NILC, since these 
parallelograms have equal bases, CH and CL, and the same 
altitude, can also be superposed by a different arrangement of 
the parts of either, as shown in Prop. 538; hence our propo- 
sition is proved. 

331. Cor, 1.—Two parallelograms, which are equal in area, 
may be superposed by two successive different arrangements of 
their parts; for first they can both be turned into rectangles, 
and then one of these rectangles superposed on the other. 

332. Cor. 2.—Two right parallelopipedons, having equal 
altitudes and bases equal in area, are equal in volume; for 
when their bases are superposed by rearrangement, the parallel- 
opipedons erected on the new bases will coincide; but their 
volumes will be equal to those they had before, having been 
formed by rearrangement of equal parts. 

The volume of each is equal to that of a rectangular parallel- 
opipedon, having the same altitude and a base equal in area. 

333. Cor. 3.—The volume of aright triangular prism is equal 
to that of a rectangular parallelopipedon having the same alti- 
tude and a base equal in area; for the volume of the right 
prism erected on the base ABC is equal 
to the volume of one of the same altitude a £ p 
erected on the equal base BCD; hence, 
either one is equal to half the volume of 
the right parallelopipedon of the same 
altitude erected on the base ABDC; but 
if we draw FE parallel to, and half way between, AB and CD, 
the volume of the right parallelopipedon of the same altitude 
erected on ABEF, or FEDGC, is also equal to half that of the one 
erected on ABDC; hence, the volume of the prism erected on 
ABC is re to that of either of the parallelopipedons erected 


98 “ELEMENTS OF GEOMETRY. 


on ABEF or FEDO; but either of these is, by Cor. 2, equal to 
that of a rectangular parallelopipedon having an equal altitude 
and a base equal in area to ABEF or FEDC, or their equal, 
ABC. 

334. Prop. 104.—The volumes of two rectangular parallelo- 
pipedons, having equal altitudes, are to each other as the wreas 
of their bases. 

Let the two parallelopipedons be denoted by P and P’, and 

A let the base of P be AB, and that of P’, 
CD. Construct a rectangle CE with an 
altitude equal to that of CD, and a base 

B equal to AB divided by this altitude ; it 
2 will (157) be equal in area to AB. Now 


construct a rectangular parallelopipedon, 
ae ahs which we will call P”, on this base CE, 
Dp £ with an altitude equal to that of P and P’. 

We shall have (127), 


vol P” _ area CE area AB 
vol P’ ~ areaCD ~ area OD’ 
But (832), vol P’ = vol P, since area CE = area AB; 
hence we have, 








vol P-" ~area ACB 
vol P’ ~ area CD’ | 
335. Cor. 1—If we now construct another rectangular 
parallelopipedon P’’”’, having a base equal to that of P’, that 
is, equal to CD, but a different altitude, we shall (127) have, 
vol P’ alt P’ 


vole oe altelas 
Multiplying this equation by the one last obtained, we have, 
since alt\P’ =—\alt Py 
vol P. «alt P area AB 


vol P” — alt P” ~*~ area CD’ 
or, expressing the altitudes in terms of any one unit of linear 
measure, and the areas of the bases in terms of any one unit of 
superficial measure, 

vol P __ alt P= X area AB, 

voL.P"): Stalt Py area” 
or, the volumes of any two rectangular parallelopipedons are 








ELEMENTS OF GEOMETRY. 99 


to each other as the products of theur altitudes and the areas 
of their bases, the former being expressed in terms of any one 
unit of linear measure, and the areas of their bases in terms 
of any one unit of superficial measure. 

336. Cor. 2—The square of the linear unit being taken as 
the superficial unit, and the cube, whose base is the square of 
the linear unit (or the cube of the linear unit as it is called), 
being taken as the unit of volume, and putting this for P’” in 
the last equation, we have 


WOlEe ia ltye ee Alea. A Io's 


or, the volume of any rectangular parallelopipedon is equal to 
the product of its altitude and the area of its base, the proper 
units being adopted for the altitude, base, and volume. 

337. Cor, 3.—The same is true (832 and 333) for any right 
parallelopipedon or right triangular prism. 

338. Prop. 105.— The volume of any prism rs equal to the 
product of tts altitude and the area of tts base. 

This is true (3837) for any right prism ; for every right prism 
can be separated, by diagonal planes passed through its edges, 
into right triangular prisms. ! 

But every oblique prism may be conceived as made up of an 
infinite number of right prisms of the same base that it has, 
and of infinitely small altitude, piled obliquely on each other ; 
the sum of the volumes of these prisms will be equal to that of 
the oblique prism; but the sum of their volumes is equal to 
the product of the sum of their altitudes by the area of their 
base; and the sum of their altitudes is equal to the altitude of 
the oblique prism, and their base is the same as its base; 
hence, the volume of an oblique prism, as well as that of a 
right prism, is equal to the product of its altitude and the area 
of its base. 

_ 839. Cor. 1—The volume of a prism may be shown in the 
same way to be equal to the product of its edge and the area 
of its perimeter; for a right prism, this is equivalent to the 
proposition just proved; and every oblique prism may be con- 
ceived as composed of a bundle of infinitely slender right 
prisms, running from one base to the other, and having an 
altitude equal to the edge of the oblique prism. The sum 


100 ELEMENTS OF GEOMETRY. 


of the volumes of these prisms is equal to the product of 
their altitude and the sum of the areas of their perimeters ; 
but the sum of these areas is equal to the area of the perime- 
ter of the oblique prism; and the sum of their volumes is 
equal to that of the oblique prism; hence, our statement is 
proved. 

340. Cor. 2.—The proposition is true of a cylinder, which is 
only a kind of right prism. 

341. Schol.—The area of a parallelogram might have been 
proved equal to that of a rectangle of the same base and _alti- 
tude, by a method similar to that employed in this proposition ; 
and this proposition might be proved by a method similar to 
the one used for that proposition, but which would be more 
complicated in this case. 

342. Prop. 106.—TZhe average square of the regular series 
of integers from 1 to any number, represented by n, mnclusive, 
as equal to $(m + 1)(n + 4). 

This is a purely algebraical proposition, but as it is not 
usually given in works on algebra, it may be as well to prove 
it here. It is only a par Hoaled case of a more general formula. 

We have, i the binomial theorem, 


= (de -b dee ee Sore, 
miCMno Mini aae Ne. 


Gat ue nN, as ae 


Adding these equations, first and third members, and noticing 
that the series 2° + 3° + etc. + 7’ is common to both members 
of the sum, and therefore dropping it, we have, | 


(vn +1 =2+1+4+3(1+2 + ete.4 n)+3(1 + 27+ ete. + n?). 


But 1 + 2 + etc.+ 2 is known by the formulas of arithmeti- 
cal progression to be equal to eats) 
this value, and transposing, we have, 
3(1 + 2? + ete. + nv’) = (n + 1) — (1 + Bn) (rn +-:1) 
= (n+ 1) (n+ 1P—1 — gn]. 
= (n + 1) [n? + gn] 
=(n +1) (m+ $)n. 


; hence, substituting 


ELEMENTS OF GEOMETRY. 101 


But the average square of the integers from 1 to m inclusive 
is equal to pana) ioe uw sea ; dividing, then, both members 
of this last equation by 38n, we have this average square equal 
to $m + 1) (x + 4), which was to be proved. 

343. Cor.—We have, gAbebass oat dna =4 (1 a ~ ) (1 oF =) : 

n n an 
and when 7 is infinite, the second member becomes simply 4. - 

Or, the average square of the regular series of integers 
from 1 to infinity is one-third of the square of the last number 
of the series. : 

344, Prop. 107.— The volume of a pyramid is equal to one- 
third of the product of its altitude and the area of its base. 

For, conceive the pyramid to be divided by a series of equi- 
distant planes parallel to its base, at the same distance from 
each other that the upper one is from the vertex, or the lower 
from the base, and let prisms be erected on each of the sec- 
tions made, and on the base, each having an altitude equal to 
the distance between the planes. 

The areas of the bases of these prisms are (313) propor- 
tional to the. square of their distance from the vertex; hence, 
representing the number of them by », the sum of their vol- 
umes is (338) equal to [1 + 2? + ete.+ n?] x the base of the 
smallest x the common altitude of the prisms; or equal to 
ae--2* + etc. + in’ 
n? 
altitude of the prisms ; 
the average square of the integers from 1 to n 

n? 
the base of the pyramid x its altitude, since the altitude of 
the pyramid is equal to 7 times the common altitude of the 
prisms. 

But if we make the common altitude of the prisms infinitely 
small, and their number infinitely great, their volume will be 
equal to that of the pyramid; and by Prop. 106, Cor., and 
the result just obtained, it will be equal to one-third of the 
product of the altitude of the pyramid and the area of its base ; 
hence, the volume of the pyramid is equal to this, which was to 
be proved. : 


x the base of the pyramid x the common 


or equal to 


102 ELEMENTS OF GEOMETRY. 


845. Cor.—This is true for a cone, which is only a kind of 
right pyramid. 

346. Schol.—It is evident that the volume of any structure 
is equal to the product of the average area of its sections made 
by parallel planes at infinitely small and equal distances, and 
the distance between the first and last of these planes; just as 
in Prop. 55, Cor. 2, the area of any plane surface is shown to 
be equal to the product of the average of its breadths at infi- 
nitely small and equal distances, and the distance between the 
first and last of these breadths. 

We may consider the volume as formed by the movement of 
the area of the section in a direction perpendicular to itself ; it 
will then be equal to the product of the whole movement of 
this area in this direction, and the average value of the area at 
infinitely small and equal distances. 

347. Prop. 108.— The volume of a frustum of a pyramid is 
equal to the sum of the volumes of three pyramids, having a 
common altitude equal to the altitude of the frustum, and 
' bases the area of which is equal respectively to that of the 
upper base of the frustum, that of the lower base, and a mean 
proportional between the two. 

For the volume of the frustum is equal to the volume of the 
pyramid of which it is a frustum, diminished by the volume of 
the pyramid which has been cut off to make it. 

Let the altitudes of these pyramids be represented by @ and 
a’ respectively, and the areas of their bases by 6 and 0’ respec- 
tively ; their volumes will (844) be $a and ja’b’ respectively, 
and the volume of the frustum will be 


4 (ab — 00) = | a (2 + 0’ + ‘a — a (2 + 0’ + vs) 
But (3138) we have, 
Oe 2. enc en af Vie b 
b ~~ a’ or yates ae a’ V pp 
and putting in these values of © and 7 the expression just 


obtained for the volume of the frustum becomes 
za—a)(64+ 0+ V bb’) ; 
which was to be proved. 


ELEMENTS OF GEOMETRY. 103 


348. Cor.—This is true for a frustum of a cone, a cone being 
only a kind of right pyramid. 

349. Prop. 109.—The volume of a sphere ws equal to one- 
third of the product of rts radius and area. 

For the volume of a sphere may be conceived as composed 
of the volumes of an infinite number of right pyramids with 
infinitely small bases lying on the surface of the sphere, and 
with a common vertex at its centre; the sum of the volumes of 
these pyramids is equal. to one-third of the product of their 
common altitude and the sum of the areas of their bases; but 
the sum of their volumes is equal to the volume of the sphere, 
their common altitude is its radius, and the sum of the areas of 
their bases is equal to the area of the sphere; hence our prop- 
osition is proved. 

It may also be proved in another way. 

Erect on the great circle of a hemisphere a cylinder with an 
altitude equal to the radius of the 
sphere, so that the upper base ‘of the 
cylinder is tangent to the hemisphere. 

Make a section of the cylinder and 
hemisphere be a plane MN between 
those of the bases of the cylinder and 
parallel to them. It will cut the 
cylinder and hemisphere in two circles, having a common centre 
at A, the foot of the perpendicular on the cutting plane from 
E, the centre of the sphere to which the hemisphere belongs. 

The area of the ring between these circles is (254) equal to 
a times the difference of the squares of their radii; but (B 
being a point on the inner circle,) the radius of the inner circle 
is AB, and that of the outer one is CB. Hence, the area of the 
ring 1s (178) 7 . CA’. 

But if we should construct an inverted cone with its base on 
the upper base of the cylinder, and its vertex at C, its section 





2 
by MN would (818) have an area equal to times the area 
2 
eet . 7. DE*, by Prop. 81; but 


DE = CD; hence the area of the section of the cone would be 
equal to 7. CA’. 





of its base, or equal to 


104 ELEMENTS OF GEOMETRY. 


Hence the volume of the cone is equal to the volume be- 
tween the cylinder and the hemisphere; or the volume of a 
hemisphere ts equal to the difference of volumes of a cylinder 
and cone having bases equal to the great circle of the hemis- 
phere, and an altitude equal to tts radius. 

But the volume of the cone is one-third of that of the cylin- 
der; hence, that of the hemisphere is two-thirds of that of the 
cylinder, or it is equal to two-thirds of the product of its radius 
and the area of a great circle, and that of the sphere is equal to 
one-third of the product of its radius and the area of four 
great circles. But the area of four great circles is (826) equal 
to that of the sphere; hence our proposition is proved. 

350. Cor. 1.—The volume of the sphere is equal to two- 
thirds that of the circumscribed cylinder, since that of the 
hemisphere has just been shown to be two-thirds of half this 
cylinder ; and the area of the sphere is also two-thirds of that 
of the circumscribed cylinder ; for the convex surface of this 
cylinder is equal, by Prop. 95 and Prop. 78, Cor. 1, to the area 
of four great circles, and its whole area is therefore equal to 
six great circles. But that of the sphere is four great circles, 
which is two-thirds of this. 

351. Cor. 2.—The parts of the volume of the sphere which 
is included between two parallel planes passed through A and 
A’ respectively, or, as it may be called, the volume of the zone 
formed by these planes, is seen by the second method of de- 
monstration to be equal to that of the portion of the cylinder 
included between these planes, diminished by the portion of the 
cone which has been constructed, included between these same 
planes. 

852. Cor. 8.—The volume of any structure of planes cir- 
cumscribed about a sphere is equal to one-third the product of 
the radius of the sphere and the area of the structure. For it 
is equal to the sum of the volumes of the pyramids having the 
planes for bases and a common vertex at the centre of the 
sphere. 

353. Def—A structure of planes enclosing a definite space, 
is called a polyedron. 

Its volume may usually be separated into those of a number 
of pyramids, having its planes for bases, and a common vertex 


ELEMENTS OF GEOMETRY. 105 


somewhere on or within the polyedron. Where this cannot be 
done, owing to some peculiar shape of the polyedron, like that 
represented for a polygon in Prop. 35, Scholium, it can be 
divided by sections into two or more polyedric volumes, for 
which this can be done. We shall not have, however, to con- 
sider such polyedrons. 

The plane areas forming the polyedron are those of..poly- 
gons, formed in each plane by the intersection of #his plane 
with the others adjacent to it. They are called the faces of the 
polyedron. The lines of intersection of the facesgor the’sides 
of the polygons which bound them, are called its edges ; the 
points where the edges meet, or the vertices of fhe polygons 
bounding the faces, are called its verteces. \ "2. ) 

The prism and pyramid are polyedrons; the tert  edods”) 
has, with them, been applied for the sake of conv@ fence, tO~ 
some only of the polyedric edges, the others not nee be to be 
referred to. See 

354. Prop. 110.—A polyedron may be constructed, eae 
faces making the same diedral angles with each other as those 
of any gwen polyedron, and bounded by polygons similar in 
any required ratio, the same for all, to those which bound the 
Saces of the given polyedron. 

From any point within the given polyedron, draw lines to all 
the vertices; and lay off on these lines, prolonged if necessary, 
distances in the ratio to their present length that the lines 
bounding the faces of the new polyedron are required to have 
to those bounding the faces of the given one. 

Pass planes through the points thus determined, parallel to 
planes of the faces of the given polyedron ; any one of these 
planes will (290) pass through all the points determined on the 
lines passing through the vertices of some one of the polygons 
of the given polyedron ; and hence they will form a new polye- 
dron, each face of which will be parallel to one of the faces 
of the given one. Its edges will also be parallel (126) to those 
of the given one. 

This polyedron will be the one required. 

For, pass a plane MN perpendicular to one of the edges AB 
of the given polyedron, prolonged if necessary. This plane 
will also be Reercndicaie to the edge CD of the new one, which 












106 ELEMENTS OF GEOMETRY. 


is parallel to AB. Now the faces of the original polyedron 
which meet at AB will cut this plane in 
lines EF and EG, and FEG will be the 
diedral angle of these planes, since EF and 
EG are perpendicular to AB. Similarly 
the faces of the new polyedron which 
meet in CD will cut the plane in lines 
HK and HL, and KHL will.be their 
diedral angle. 

But KH is parallel (279) to EF, and 
KL to EG, since the planes of the faces of the polyedrons are 
parallel to each other; hence, the diedral angles FEG and 
KHL are (188) equal; and this might be proved of any other 
two corresponding diedral angles. So, in this respect, our new 
polyedron is the one required. 

Secondly, the polygons bounding the faces of the two polye- 
drons formed by two parallel planes, will be the base and sec- 
tion of a pyramid, and they will be similar to each other, 
(311) in the ratio of the distances of their vertices from the 
point within the original polyedron from which lines were at 
the outset drawn to its vertices; but this is the required ratio; 
hence, in this respect also, our new polyedron is the one re- 
quired, and our proposition is proved. 

355. Defi—Such polyedrons are said to be semdar to each 
other. The similar faces are called corresponding faces ; the 
corresponding sides and vertices of the polygons bounding 
them are called corresponding edges and vertices of the polye- 
drons. 

Points similarly situated in the corresponding faces are 
called corresponding points; and the lines connecting them 
are called corresponding lines. 

856. Prop. 111.—The areas of two sumilar polyedrons are 
to each other as the squares of their corresponding edges ; and 
their volumes are to each other as the cubes of their corre- 
sponding edges. 

For the areas are the sums of the areas of the faces ; but the 
‘areas of the faces are to each other (181) in the ratio of the 
squares of the edges; hence, the areas of the polyedrons also 
are 80. 





ELEMENTS OF GEOMETRY. 107 


And the volumes are sums of those of pyramids having these 
faces for their faces, and altitudes proportional to the edges; 
hence (844), the volumes of the pyramids are to each other as 
the cubes of the edges, and the volumes of the polyedrons are 
in the same ratio. 

357. Cor. 1.—This is true for similar prisms and pyramids, 

they being polyedrons. 
— 858. Cor. 2.—Two polyedrons having equal faces similarly 
placed, and equal diedral angles, can evidently be superposed ; 
hence, all the polyedrons formed from the various points within 
the given polyedron of Prop. 110, with the same ratio of dis- 
tances, are identical; hence, also, we may define similar polye- 
drons as having faces similar in a uniform ratio, and similarly 
placed, and having equal diedral angles between the similar 
faces. | 

It is plain that the pyramids used in the demonstration of the 
present proposition are similar. 

359. Cor. 3.—Instead of the edges we may substitute in the 
enunciation any corresponding lines; for these will have the 
ratio to each other that the distances of their extremities from 
the point within around which the polyedrons are formed 
have, and this ratio is that of the edges. 

360. The cylinder, cone, and sphere are sometimes called the 
three vound bodies. The following formulas give their areas 
and volumes, denoting by R the radius of the sphere and of 
the bases of the cylinder and cone, by A the altitudes of the 
cylinder and cone, and by D the diameter of the sphere: 


Convex area of cylinder = altitude x base = A x 27R = 27AR; 


eee ee he rr Zar he = 2 Ab Th) Th 

Convex area of cone = altitude x base = 7wAR; 

Titel ce Oh a = wAR + wR? = w(A + RB) R. 

Area of sphere = area of four great circles = 47K 

earl yc. 
Volume of cylinder =altitude x area of base = Ax7h* 
me ya atap | 

eee. cone = 4 altitude x area of base = 47AR’; 
ame sphere = + radius x area of sphere = 47h’. 


To these we may add the areas and volumes of the frustum 


108 ELEMENTS OF GEOMETRY. 


of a cone, and of a spherical zone or segment, (which may be 
called the frustum of a sphere ;) denoting by R’ the radius of 
the upper base of the frustum of the cone; and by A the alti- 
tude of the frustum, zone, or segment; and by P and P’ the 
distances of the planes forming the zone or segment from the 
centre of the sphere : 


Convex area of frustum of cone = altitude x 4 sum of bases 
Whole “ & % ec  &€ =arA(R + BR) 4+ cR’+7R? 
=m{[At+R)RF+(A 


+ RRA: 
Area of zone or segment —Ax2rR=27AR=7AD. 
Volume of frustum of cone = trA (R? + R? + RR); 
« © zone or segment (851) = wh? (P += P’) — 47 (P8 
+P"); 


the upper or lower signs being used according as P and P’ 
are on the same side or on opposite sides; or this volume is 
equal, since P = P’= A, to 


TA [| R*— 4b? 4 Pea 
the upper or .ower signs being used as before. The formula 


for the volume of the frustum of a cone may also be written 


A 3 13 
Tipe ee 


BOOK IX. 
SPHERIOAL GEOMETRY. 


361. Def—An indefinite pyramid, that is, a pyramid the 
sides of which are indefinitely prolonged, and which has no 
base, is called a polyedral angle. 

It is composed simply of the planes meeting in the vertex 
and edges of the pyramid. 

These planes are called its sedes; the angles made by the 
adjacent edges its plane angles, and the angles made by the 
planes its diedral angles. The vertex of the pyramid is called 
its vertex. | 

A polyedral angle of three sides is called a t¢riedral angle. 
The triedral angle is the principal subject of spherical geometry. 

The diedral angle included between the planes of two great 
circles of a sphere, is called the spherical angle between those 
circles. 

The circles are called the sedes of the angle; the point where 
they meet is called its vertem. 

The circles will meet also at the opposite end of the diameter 
in which their planes intersect ; the two semicircles running 
from one point of. meeting to the other, form what is called a 
lune. The angles at the two ends of the lune are the same. 

If one of the semicircles is completed, that is, made a circle, 
two lunes are formed, the angles of which are adjacent, and 
consequently have a sum equal (271) to 2h. 

If both semicircles are completed, four lunes are formed, 
each of the two new ones being equal to the one already formed 
opposite to it. These lunes will be in two pairs, each member 
of either pair having its angle opposite to that of the other 
member. Each of the pairs of lunes is called a double lune. 
(See figure of Prop. 118.) 


110 ELEMENTS OF GEOMETRY. 


362. Prop. 112.—The spherical angle between two ares of 
great circles, is equal to that between the tangents to those ares 
at their point of meeting, and it is measured by the arc of a 
great circle connecting two points on them a quadrant distant 
Jrom their point of meeting. 

Let DAE be the spherical angle considered, the sides of 
which meet at A and B, forming the lune ADBE. 

The spherical angle, being the die- 

we dral angle of the planes which form it, 

is equal to the angle made by any two 
perpendiculars in those planes to their 
M "intersection, at the same point; but the 


tangents AF’ and AG are two such per- 
iy pendiculars, hence the spherical angle 
4 is equal to the angle FAG. 


If we pass a plane perpendicular to 
the diameter AB at the centre, cutting the sphere in the great 
circle MKL; all radii as CK and CL in this plane will be per- 
pendicular to AB, and hence the spherical angle is equal to the 
KCL for the same reason that it was equal to FAG; but CK 
and CL being perpendicular to AB, the ares ADK, AEL, are 
quadrants, hence KL is the arc of a great circle spoken of in 
the second part of the enunciation. But KCL is measured (193) 
on the great circle MKL by KL; hence our proposition is 
proved. 

363. Schol—tThere is a certain pr opriety in calling the 
diedral angle of the planes of two great circles their spherical 
angle, oy in considering it as the angle which the arcs them- 
selves make with each other; since it is the angle which their 
instantaneous directions at ie point of meeting, indicated by 
their tangents, make with each other. 

364. Def—The figure, (such as KAL,) formed on the sur- 
face of a sphere by three plane faces, (such as ACK, ACL and 
KCL,) forming a triedral angle with its vertex at the centre of 
the sphere, is called a spherical triangle. 

It has three sides, and three spherical angles. The sides are 
each less than a semicircle, and the pele are each less than 
two right angles. | 

‘The arcs, which are naturally called its sides, are proportional 


ELEMENTS OF GEOMETRY. 111 


to the plane angles forming the triedral angle, or measure 
them; but the sides, properly so called, of the triangle are the 
plane angles of the triedral angle, and the angles of the triangle 
are the diedral angles of the triedral angle. 

The first distinction is specially important; for if the ares 
were really the sides of the triangle, the triangle would be 
larger or smaller, according as the sphere on which it was con- 
structed were larger or smaller; but the spherical triangle, 
properly so called, and all other figures drawn in spherical 
geometry, are not dependent on the size of the sphere at all. 

The spherical triangle is really only a triedral angle; the 
sphere is only brought in in order that we may more easily 
discuss its properties. 

It is true, however, that we may sometimes have an actual 
spherical triangle to consider, on some definite sphere; in this 
case, we must take account of the size of the sphere, which can 
easily be done. The spherical triangle in the figure has two 
right angles ; but this is not necessarily the case, for the planes 
forming it may make any triedral angle whatever. 

_The planes making the triedral angle produced in every di- 
rection, will make eight spherical triangles, as will be seen on 
consideration ; habitually, however, we consider only one. 

A polyedral angle in general, with its vertex at the centre of 
the sphere, forms what is called a spherical polygon. 

As a triedral angle is only a kind of polyedral angle, so a 
spherical triangle is only a kind of spherical polygon. 

The diagonal of a spherical polygon is an are of a great 
circle connecting two vertices not adjacent. 

365. Prop. 1138.—Any one of the sides of a spherical trv- 
angle is less than the sum of the other two, 


and greater than their difference. K 
Let V be the vertex of the triedral angle 
which forms the spherical triangle, and let 
AVC be a plane angle in it which is larger 
than either of the others; it is only fora , + F 


side of the triangle measuring such an an- 
gle that the first part of the proposition re- f 
quires proof. 

Draw a line AC at pleasure connecting the sides of this 


112 ELEMENTS OF GEOMETRY. 


plane angle, and lay off a point D on it such that the angles 
AVD and AVB are equal. Then lay off on VB a distance 
Vi = VD, and draw AE and CE. 

We have (28) AD = AE; but (42), AC < AE + EO; 
hence AC — AD < EC, or DC < EC; hence (47) the angle 
DVC is less than the angle BVYC; hence AVD + DVO < 
AVB + BVO, or AVC < AVB + BVC. Hence, the side 
of the spherical triangle measuring AVC is less than the sum 
of the other two, as was to be proved. 

The second part of the proposition results immediately from 
the first, as in Prop. 14. 

366. Cor.—The are of a great circle is the shortest line 
which can be drawn on a sphere between two points on that 
sphere; or in other words, a line undergoes less rotation m 
passing from one position to another intersecting the first 
of it is turned around a fixed axis perpendicular to the plane 
of the two positions, than if a different or variable axis be 
taken. 

This is proved just as Prop. 15 is proved from Prop. 14. 

3867. Def—The extremities of the diameter perpendicular - 
to the plane of any great circle, are called the poles of that 
circle. Thus A and 5, in the figure of Prop. 112, are the 
poles of the circle MKL. 

They are also called the poles of all the small circles formed 
by planes parallel to that of the great circle. 

All great circles passing through the poles of a great circle 
are (273) perpendicular to that great circle. Their planes are 
also, by the same proposition, perpendicular to those of the 
small circles. } 

The arc between either pole of a great circle and any point 
of that circle is a quadrant, since the angle which it measures 
is a right angle. 

The arcs between either pole of a small circle and all the 
points of that circle (197), are equal; for the chords of these 
arcs are hypothenuses of right triangles which are (28) equal. 
These ares are less or greater than a quadrant, according as the 
pole from which they are drawn is the one nearer to, or farther 
from, the plane of the small circle. 

368. Prop. 114.—Hrom a point outside of a great curcle 


ELEMENTS OF GEOMETRY. 113 


and not its pole, one great circle, and only one, can be drawn 
perpendicular to that circle. 

For drop from this point an axis or perpendicular to the 
plane of that circle. A plane including this axis and the 
centre of the sphere will (278) be perpendicular to the plane 
of the circle, and consequently will form on the sphere a great 
circle perpendicular to that circle. 

No other great circle can be drawn perpendicular to that 
circle through this point; for if so, its plane must be perpen- 
dicular to the plane of that circle, and hence, must include by 
the second part of Prop. 88, the axis or perpendicular which 
has been drawn; and it must also pass through the centre of 
the sphere, otherwise the circle which it forms would not be a 
great circle. 

But only one plane can be passed so as to include this axis 
and the centre of the sphere, namely, the one already passed. — 

369. Cor. 1—The plane of the one great circle perpen- 
dicular to the given one, since it passes through the centre of 
the sphere, includes (273) the diameter perpendicular to the 
plane of the given one; hence, the great circle which it forms 
passes through the poles of the given circle. 

370. Cor. 2.—Two great circles which are perpendicular to 
a third, intersect at the poles of that third, as is clear from the 
last corollary. 

371. Cor. 3.—lf the distance of a point from two points of 
a great circle is equal to a quadrant, that point is a pole of the 
circle. For the radius of the sphere drawn to that point must 
be perpendicular to the radii drawn to those two points; hence 
it is the axis of the only (20) plane, including those radii, and 
its extremity is therefore a pole of the circle formed by that 
plane. 

372. Schol—A great circle, or any are of it, may be de- 
scribed around its pole as a centre, with a circular radius equal 
to a quadrant, in the same way as a circle can be described in 
a plane. 

373. Prov. 115.—J/f from the vertices of the angles of a 
spherical triangle as poles, great circles be drawn, the vertices 
of the eight triangles which these circles will form, will be poles 
of the sides of the original triangle. 


114 ELEMENTS OF GEOMETRY. 


Eight triangles will be formed, as will be seen on inspection 
of the figure of Prop. 118; but they will only have six vertices, 
which will be the extremities of the intersections of their three 
planes, and these six vertices will be in three pairs, each mem- 
ber of each pair being opposite to the other one. 

Let ABC be a spherical triangle ; and aa, 6b, cc, semicircles 
described from A, B, C, respectively, 
as poles; and let A’, B’, C’, be the 
points in which 6d and cc, ce and aa, 
aa and 6b, respectively, intersect on 
this side of the sphere. There will be 
three more points on the other side 
opposite to these respectively. 

Now A’, being on 00, is distant a 
quadrant from B; and, being on ce, | 
it is distant a quadrant from C; hence (871), it is a pole of 
BC. Similarly, B’ and C’ are poles of AC and AB. 

The points opposite A’, B’, and C’, will of course also be 
poles of BC, AC, and AB, respectively, and of the sides of the 
seven other triangles formed by their circles, as A, B, and C, 
are poles of B’O’, A’C’, and A’B’, and of the sides of the seven 
other triangles formed by their circles. 

374. Def—Two sets of triangles thus related are said to be 
polar to each other. 

375. Schol.—F rom the opposition existing in the six vertices 
in each set of triangles, it is plain that the arc between two 
members of two pairs will be the same as between the opposite 
members of the same two pairs, if we go around the circle on 
which these four vertices he in the same direction in both 
cases; and hence that the opposite triangles formed by three 
great circles have their sides equal, each to each. The eight 
triangles of each set may then be divided into two sets of four 
each, the one set being equilateral to the other set, but the 
members of each set differing among themselves. 

If the set of four be made up of one triangle, as A’B’C’ in 
the figure, and three others, each of which has one common 
side with it, as the uncompleted triangles cB’C’}, aO0’A’e, 
6A’B’a, of the figure, and if the semicircle be denoted by z, 
(as is usually done, it being a times the radius,) the sides of the 





ELEMENTS OF GEOMETRY. 115 


three other triangles will be B’O’, 7 — A’C’, 7 — A’B’; 
a — BO’, A’C’, 7 — A’B’; and a — BC", 7 — A'O, AB, 
respectively. 

376. Prop. 116.—I/n the figure of the last proposition, the 
angles A, B, C, are measured by r—B'C'", r—A'C'", w— A'B’, 
respectively, and A’, B’, C, byrw— BC, 7 —-AC, 7 — AB, 
respectwely. 

For produce the sides AB, AC, for instance, till they cut 
B’C’ in two points B”’, C”. A is the pole of B’C’, hence AB” 
and AO” are quadrants, and hence (362) the angle A is meas- 
Greve ©. (Bat.B’C"= BC’ + BC — BC’; and B/C” 
and B’C are each equal to a quadrant or $7, since B’ and C’ 
are (373) the poles of AC and AB respectively. Hence A is 
measured by 7 — B’C’, as was to be proved. The same could 
be shown for the other angles. 

Similarly it can be shown that the angles A’, B’, CO’, are 
measured by 7 — BO, 7 — AO, 7 — AB, respectively. 

377. Schol. 1—In case B’ and OC” do not fall between 
B’ and ©’, we should have either C’B” + B’C” = O”’B’ 
+ B’O,, or, B’C” + C”B’ = B’C’+ C’B”; or in either case 
B'C’ = B”C”, since the other arcs are quadrants; the angle 
A would then, in this case, not be measured by 7 — B’O, 
but by B’C’ itself. In this case, not A’B’C’, but one of the 
other triangles belonging to its set, having a side equal to 
a — B/C’, will (875) be the one for which the proposition 
will hold. This triangle, whichever it is, is the one properly 
called the polar triangle of ABC. | 

3878. Schol. 2.— ABC is also polar to its polar triangle prop- 
erly so called. That this may be seen more clearly, let us con- 
sider the extreme forms which a spherical triangle may assume, 
viz., a point and a great circle. The great circle is polar to the 
point, and the point to the circle. Let us now continuously 
increase the sides of the small triangle conceived as condensed 
at the point. In the beginning this triangle and one of its 
polar set can be made to have the relation to each other of the 
figure of ABC and A’B’C’, B” and C” both falling between 
B’ and C’, and so for the rest ; and the triangles will be strictly 
polar, the one to the other. As we expand the small triangle 
by any continuous changes of its sides up to a semicircle, its 


116 ELEMENTS OF GEOMETRY. 


polar triangle will diminish continuously to a point, but the 

triangles must, by the general principle of continuity, always 

retain their mutual polarity. 

Every spherical triangle, then, has its polar, with sides equal 
to a semicircle diminished by the measures of the angles of the 
first, and angles measured by 2R diminished by the angles 
measured by the sides of the first. The triangles need not be 
considered as actually placed in the position of polarity; the 
polar to any triangle can be drawn on any part of the sphere ; 
or, to use more comprehensive terms, the triedral angle polar 
to a given triedral angle, can be located anywhere in space. 

379. Prop. 117.—Two spherical triangles having the three 
sides of the one equal to the three sides of the other, are equal 
in all their parts. 

Let V, V’ be the vertices of the triedral angles which form 

the two triangles. 

v From any two points A and A’ equally 
distant from V and V’ on edges between 
any two of the equal plane angles which 
form the equal sides of the triangles, | 
draw lines AB, AC, and A’B’, A’O’, in 

,; ¢’ the planes of the triedral angles, and 

we perpendicular to VA and V’A’ respect- 
ively. 
The triangles VAB and V’A’B’ are 

(51) equal; hence, A’B’ = AB, and V’B’ = VB. 

In the same way it can be shown that A’C’ = AC, and V’C’ 
= VC. Hence, the triangles VBC and V’B’C’ are (28) equal, 
and hence (48) the triangles ABC and A’B’C’ are also equal ; 
hence the angles BAC and B’A'C’ are equal. 

But these angles are the diedral angles, or the angles of the 
spherical triangle, since the lines AB, AC, and A’B’, A’C’%, are 
perpendicular to VA and V’A’ respectively; hence, these 
angles of the triangles are equal, and the same could be shown 
for the other angles; hence, our proposition is proved. 

380. Cor.— Two spherical triangles having the three angles 
of the one equal to the three angles of the other, are equal m 
all their parts. For their polar triangles must have the sides 
of the one equal to those of the other, by Prop. 116; hence, 





ELEMENTS OF GEOMETRY. Ti? 


these polar triangles have the angles of the one equal to those 
of the other, by this proposition; hence, again by Prop. 116, 
the original triangles have the sides of the one equal to those 
of the other. 

881. Schol.—The equal sides may not come in the same 
order going around the two triangles in the same direction on 
the sphere ; they still have equal angles, but are not capable of 
superposition, since they cannot, on account of the curvature 
of the sphere, be reversed like plane triangles. (Prop. 8, Schol.) 

Such is the case with the opposite triangles formed by the 
same planes on a sphere, since their sides are in the same order 
if we look in the same direction along the double triedral angle 
forming them, that is, if we look at one from the outside. the 
other from the inside of the sphere, and conse- 
quently in an opposite order if we look at both 
from the outside or from the inside. 

382. Def—Such triangles are called symme- 
trical triangles. They are so called because they 
can be placed symmetrically, asin the figure, on 
the opposite sides of the same great circle. 

383. Prop. 118.—Symmetrical triangles are equal im area. 

For, place them in a position of opposition to each other, the 
edges of the triedral angle of one being prolongations of those 
of the other (Prop. 117, Schol.). Fix a number of points at 
pleasure within the spherical area of either one, and draw 
straight lines connecting them with each other and with the 
sides of the triangle, and the points on the sides of the triangle 
with each other, thus forming a network of plane triangles. 
Draw radii from the vertices of these triangles to the centre of 
the sphere, and produce them to the opposite surfaces; and 
connect their terminal points by straight lines, forming a simi- 
lar network of plane triangles in the area of the symmetrical 
triangle. Each side of any one of these plane triangles in 
either set will be equal (28) to the corresponding side on the 
opposite surface; hence, each triangle will be equal (48) to the 
opposite triangle; and hence, also, equal in area. Hence, the 
area of the polyedric surface formed on one side of the sphere 
will be equal to that formed on the other side. But if we take 
an infinite number of points, making triangles, the sides of 


118 ELEMENTS OF GEOMETRY. 


which are infinitely small, the polyedric surfaces will coincide 
with those of the symmetrical triangles; hence, our proposition 
is proved. 

384. Cor.—The two triangles ABC, A’bB’C, formed on the 
same hemisphere by two intersecting 
circles, have together an area equal to 
that of the lune CAO’B, having an 
angle C equal to the common angle of 
the triangles. Jor the area of the lune 
is equal to the sum of the areas of 
ABC and ABC’; but the area of ABO’ 
is, by this proposition, equal to that of 
A'B'C. 

385. Defi—If a lune is cut by a great circle having the ex- 
tremities of the lune as poles, the two equal triangles formed, 
are called 62-rectangular triangles. They have two right an- 
gles, the third angle being that of the lune; the sides opposite 
the right angles are quadrants; the third side measures the 
angle of the lune. 

If the angle of the lune is a right angle, the triangles have 
three right angles, and are called ¢rd-rectangular triangles. 
Their sides are quadrants. | 

The area of the tri-rectangular triangle is one-eighth of the 
area of the sphere. 

The area of a lune is evidently (127) the same fraction of 
the area of the sphere, as its angle is of four right angles. 

That of a bi-rectangular triangle is half this. 

886. Prop. 119.—The sum of the angles of a spherical tri- 
angle exceeds two right angles by an amount which is to one 
roght angle as the area of the triangle is to that of the tri- 

rectangular triangle. . 

Let ABC be aspherical triangle. Draw 
a great circle, so that its plane will not 
cut that of the three vertices A, B, C, in- 
side the sphere. Prolong the sides AB, 
BO, and AC at both ends, till they meet 
this great circle, forming three pairs of 
triangles ‘like those of Prop. 118, Cor., 
having common angles at A, B, and C, respectively. 








ELEMENTS OF GEOMETRY. 119 


The sum of the areas of these three pairs of triangles is by 


Prop. 118, Cor., equal to aS 


but it exceeds the area of the hemisphere by twice the area of 
the triangle ABC, since this has appeared in all the pairs. 
Hence, 


x the area of the sphere ; 


ee _ 5) x the area of the sphere = twice the 


area of ABO; 


ee A+B+4+C—2R | the area of ABC 

; R ~ 4of the area of the sphere’ 
hence, since the area of the tri-rectangular triangle is one-eighth 
of that of the sphere, our proposition is proved. 

387. Cor.—The sum of the angles of a spherical polygon of 
m sides, exceed 2(7 — 2) right angles by an amount which is to 
one right angle as the area of the polygon is to that of the tri- 
rectangular triangle, since m — 2 triangles can be formed in 
the polygon by diagonals drawn from any one vertex, and the 
sum of the angles of the polygon is equal to that of the angles 
of the triangles. 

The greatest possible amount of the excess of the sum of the 
angles of a spherical triangle or polygon over two right angles, 
or 2(n — 2) right angles respectively, is four right angles, since 
the triangle or polygon cannot have an area greater than that 
of a hemisphere, which is equal to four tri-rectangular triangles. 
The greatest possible sum of the angles of a spherical triangle 
or polygon is then six right angles, or [2(n — 2) + 4 = 2n] 
right angles respectively. 

388. Def—The excess of the sum of the angles of a spherical 
triangle or polygon over two right angles, or over 2(m — 2) right 
angles, respectively, is called its spherical excess, being the ex- 
cess due to its being a spherical, instead of a plane, triangle or 
polygon. 

389. Prop. 120—The sum of the sides of any spherical 
polygon, vs less than a great circle. 

For, make a section of the polyedral angle forming the 


120 ELEMENTS OF GEOMETRY. 


spherical polygon by any plane, thus forming a oe the 
base of which is ABCDE. 

From any point F of this plane, within ABCDE, draw FA, 
FB, FC, FD, and FE, thus forming a num- 
ber of Eeanoles eanel to those the common 
vertex of which is at V. 

Now (365) FAE + FAB < VAE + 
VAB; and the same will be the case at 
all the other points B,C, D, E. We have, 
then, the sum of the angles at the bases of 
the triangles around fF’, less than the sum 
of the angles at the bases of those around Y. 

But there are just as many triangles in 
both sets; hence (87), the sum of the angles 
AV B, etce., around. V, must be less than the sum of the angles 
AFB, ete., around I’, or be less than four right angles ; hence, 
the sum of the sides of the polygon, which measures the sum 
of these angles, is less than a great circle, which was to be 
proved. 

390. Def—If a regular plane polygon be inscribed in a 
small circle, and the vertices connected by ares of great circles, 
the spherical polygon formed is said to be regular, having, 
like the regular plane polygon, equal sides and equal angles. 

391. Prop. 121.— Zhe ue of a sphere cannot be divided 
into more than five sets of equal regular spherical polygonal 
surfaces. 

No set of regular polygonal areas can exactly cover the 
sphere, unless the angle of each polygon is an aliquot part of 
four right angles. 

Now the only aliquot parts of four right angles possible for 
the angle of a regular (or equiangular) spherical triangle are 
$, +, and 4, since a smaller fraction would give no spherical ex- 
cess; the only aliquot part possible for a regular spherical 
quadrilateral or pentagon is $, for the same reason, since the 
sum of the angles of the quadrilateral must exceed four right 
angles, and that of the angles of the pentagon must exceed six. 

There is no aliqnot part possible for a hexagon, or polygon 
of still more sides, for even $ gives no spherical excess for 
them. 


V 





ELEMENTS OF GEOMETRY. ABE 


392. Cor.—The regular spherical triangles having an angle 


4R 4R 4h 


equal to P apingee ea) respectively, will have spherical excesses 


of 2K, R, and a respectively, and areas equal to 4, 4, and 1, 
of that of the sphere respectively. The regular quadrilateral and 
4h. 
pentagon having an angle equal to 3 will have spherical ex- 
2 
cesses of = and 2 respectively, and areas equal to + and 7, 
of that of the sphere respectively. 

393. Schol.—If the vertices of the above sets of regular 
polygons be joined by straight lines, the planes of these lines 
will form what are called the jive regular polyedrons ; viz., 
the tetraedron, the octaedron, and the icosaedron, having tri- 
angular faces; the cube having square faces; and the dode- 
caedron having pentagonal faces. 


GENERAL COROLLARY. 


394. Propositions, similar to Props. 7 and 19, can be proved 
for spherical triangles; for if the triangles are not directly 
capable of superposition, one can be superposed on a triangle 
symmetrical to the other. 

From these propositions, others similar to Props. 8, 20, and 
21, can be deduced. 

The single great semicircle, which can, by Prop. 114, be 
drawn perpendicular to a great circle from a point outside of 
it and not its pole, has no point on its shorter part between 
this one and the circle from which another perpendicular can 
be dropped; hence, a proposition similar to Prop. 12 can be 
proved, understanding by the perpendicular in the enunciation 
the shorter of the two perpendiculars forming the semicircle. 
The corollaries of Prop. 12 will not, however, be true on the 
sphere. 

By means of the proposition similar to Prop. 21, a propo- 
sition similar to Prop. 18 readily results. 


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PAS Dee exes, 


THE following additional propositions are appended as exercises to be proved 
by the student. 

EXERCISE 1. A broken line, having its concavity always toward the straight 
line connecting its extremities, is less than any other line, broken or curved, 
which can be drawn between the same extremities, on the same side of the 
straight line with the broken one, and on the opposite side of the broken line 
from the straight one. 

A line drawn thus is said to envelop the broken line. 

Ex. 2. The three perpendiculars dropped from the vertices of the angles of 
a triangle on the opposite sides meet in the same point. 

Ex. 3. The three lines drawn from the vertices of the angles of a triangle 
to the middle points of the opposite sides, meet in the same point; and this 
point is twice as far from each vertex as from the middle point of the oppo- 
site side. It is called the medial point of the triangle. 

Ex. 4. If four lines be drawn, joining the middle points of the siiieent 
sides of any quadrilateral, they will form a parallelogram; and the perimeter 
of this parallelogram will be equal to the sum of the diagonals of the quadri- 
lateral. 

Ex. 5. If two lines be drawn, joining the middle points of the opposite sides 
of any quadrilateral, their intersection will be the middle point of the line 
joining the middie points of the diagonals of the quadrilateral. 

Ex. 6. If four lines be drawn, bisecting the four angles of a quadrilateral, 
they will form a second quadrilateral, the sum of the opposite angles of which 
will be equal to two right angles ; and if the first quadrilateral is a parallelo- 
gram, the second will be a rectangle ; if the first is a rectangle, the second will 
be a square. 

Ex. 7. A triangle can always be constructed, having for its sides the three 
lines of Ex. 2; and the area of this triangle will be three-fourths of the area 
of the original triangle for which the three lines are drawn. 

Ex. 8. The area of the parallelogram of Ex. 4, is half that of the quadri- 
lateral. 

Ex. 9. If through the middle point of each diagonal of a quadrilateral a 
parallel is drawn to the other diagonal, and from the point where these parallels 
intersect lines be drawn to the middle points of the four sides, the four quad- 
r.laterals formed by these lines and the sides will be equal in area. 

Ex. 10. (Theorem of Pappus.) If upon two sides of any triangle parallelo- 
grams be constructed of any magnitude and form, and the sides of these which 
are parallel to those of the triangle be produced till they meet; and if this 
point of meeting be connected with that of the two sides of the triangle ; and 
if, finally, a parallelogram be constructed on the third side of the triangle 


124 APPENDIX. 


having the sides adjacent to this third side equal and parallel to this connect- 
ing line; the area of this parallelogram will be equal to the sum of the areas 
of the other two. 

Prove Prop. 59 by means of this. 

Ex. 11. If the two pairs of opposite sides in a quadrilateral, inscribed in a 
circle, be produced till they meet, and lines be drawn trom the points of meet- 
ing bisecting the angles formed there, these lines will intersect at right angles. 

Ex. 12. If a circle be circumscribed about a triangle, and a chord be drawn 
cutting two of the sides so that the vertex of the angle included between 
those sides shall be in the centre of the arc which the chord subtends, the 
quadrilateral formed by the chord and triangle can be inscribed in a circle. 

Ex. 13. If from one extremity of any arca secant be drawn, meeting the 
radius to the other extremity produced beyond the centre, at such a point 
that the part of this secant between this point and the circle shall be equal to 
the radius of the circle, the angle made at the point will be one-third of the 
angle at the centre corresponding to the are. 

Ex. 14. If a circle be circumscribed about an equilateral triangle, the line 
drawn from any point of this circle to the vertex of the angle formed by the 
sides between which the point lies, will be equal to the sum of the lines from 
the point to the vertices of the other two angles of the triangle. 

Ex. 15. If two tangents are drawn toa circle from the same point, and a 
third tangent be drawn connecting points of these two which lie between their 
point of meeting and their respective points of contact; the perimeter of the 
triangle, formed by this third tangent with the other two, will be equal to 
twice that part of either of the other two which lies between its point of 
meeting with the other and its point of contact ; and the angle made by two 
lines from the centre of the circle to the points of meeting of the third tan- 
gent with the other two, will be equal to one right angle minus half the angle 
made by those two tangents with each other. 

Ex. 16. If two secants are drawn through the point of contact of two cir- 
cles, the chords of the arcs intercepted by these secants are parallel, and the 
secants are divided proportionally. 

Ex. 17. If two circles are tangent internally, and a chord of the larger circle 
te drawn tangent to the smaller one; a line drawn from the point of contact 
of the circles to the point of contact of this chord, will bisect the angle formed 
by two lines from the point of contact of the circles to the extremities of the 
chord. 

Ex. 18. If through one of the points of intersection of two circles diameters 
be drawn to both, the line connecting the other ends of these diameters will 
pass through the other intersection, and will be longer than any other line 
passing through it and terminated by the circles. 

Ex. 19. The feet of the perpendiculars let fall on the sides of a triangle 
from any point of the circumscribed circle, are in a straight line. 

Ex. 20. If two lines be drawn from the vertex of any angle of a triangle, 
within that angle, one making the same angle with one of the sides that the 
other does with the other side, and if one of these lines terminates in the 
third side, the other in the circumscribed circle, the rectangle of these lines 
is equal to that of the two sides including the angle. 


APPENDIX. 125 


If one of these lines is the perpendicular on the third side, what will the 
other be? 

Ex. 21. In any triangle, if a line be drawn bisecting one of its angles and 
terminating in the opposite side, the product of the sides including the angle 
will be equal to the square of this bisecting line plus the product of the parts 
into which it divides the third side. 

Ex. 22. If on a finite straight line a point be taken at pleasure, and another 
point be fixed on the same line, such that its distance from one end of the 
line shall be a mean proportional between the whole line and the distance of 
the first point from the same end; and if from that same end another line 
equal to this mean proportional be laid off in any direction ; and if, from the 
extremity of this line, lines be drawn to the other end of the original line and 
to the two points ; the line drawn to the second point will bisect the angle 
formed by the other two. 

Ex. 23. In any triangle, the centre of the circumscribed circle, and the 
points of meeting shown to exist in Exs. 2 and 3, are in the same straight 
line ; and the second point is twice as far as the first from the third. 

Ex. 24. In any quadrilateral inscribed in a circle, the product of the diago- 
nals is equal to the sum of the products of the opposite sides. 

Ex. 25. The perpendicular from any point of a circle on a chord is a mean 
proportional between the perpendiculars from the same point on the tangents 
drawn at the ends of the chord, 

Ex. 26. If two circles are tangent externally, the part of their common 
tangent which is included between the points of contact is a mean propor- 
tional between the diameters of the circles, 

Ex. 27. If a fixed circle is cut by any circle which passes through two 
fixed points, the common chord passes through a fixed point. 

Ex. 28. If through any point within a triangle lines be drawn from the 
three angles to cut the opposite sides, the product of either three of the six 
parts formed, which are not adjacent to each other, will be equal to the pro- 
duct of the other three. 

Ex. 29. Conversely, when the sides of a triangle are thus divided, lines 
drawn to the points of division from the opposite angles will pass through the 
same point. 

Ex. 30. If a straight line be drawn to cut the three sides of a triangle, one 
of the sides being produced, or, all if necessary, thus dividing them into six 
parts (a prolonged side being taken as one part, and its prolongation as an- 
other), then will the product of any three parts, which are not conterminous, 
be equal to the product of the other three. 

Ex. 31. Conversely, if the sides of a triangle be thus divided, the three 
points of division will lie in the same straight line. 

Ex. 32. If a quadrilateral be divided into two quadrilaterals by a line con- 
necting its opposite sides, the points of intersection of the diagonal of the 
original one, and of the two new ones will lie in the same straight line. 

Ex. 33. If in each of the sides of a triangle points be taken at pleasure, and 
circles be described through each angle of the triangle, and the points taken in 
the two sides, including that angle, these three circles will pass through a 
common point. 


126 APPENDIX. 


Ex. 84. The triangle formed by joining the centres of these three circles 
will be similar to the original triangle. 

Ex. 35. If a line be drawn passing through one of the points of intersection 
of any two of these circles (not the common point of intersection of the three) 
and terminating in the circles, and two other lines be drawn from its termi- 
nations through the other two points of intersection of two of the circles, these 
lines will meet in the third circle; and the triangle formed by these lines will 
be similar to the original triangle. 

Ex. 36. If equilateral triangles be constructed on the three sides of any 
triangle, outside of that triangle, the circles circumscribed about these tri- 
angles will pass through a common point; the lines connecting the angles of 
the original triangle with the outside angles of the equilateral triangles con- 
structed on the opposite sides are equal, and pass through this common point ; 
and the lines joining the centres of the equilateral triangles will form another 
equilateral triangle. 

Ex. 87. In any quadrilateral circumscribed about a circle, the sum of 
either two opposite sides is equal to the sum of the other two opposite sides; 
and conversely, if this is the case in a quadrilateral, a circle may be inscribed 
in it. 

Ex. 38. The area of a regular inscribed dodecagon is equal to three times 
the square of the radius. 

Ex. 39. The apothem of a regular polygon of an even number of sides 
greater than four is equal to the arithmetical mean of, and its radius is equal 
to a mean proportional between, the apothem and radius of a regular polygon 
of half the number of sides and an equal perimeter. 

This proposition furnishes us with another method of computing the value 
of z. ; 

Ex. 40. The area of a circle is a mean proportional between the areas of 
any two similar polygons, one circumscribed about the circle, and the other 
having the same perimeter as the circle. (Galileo’s Theorem. ) 

Ex. 41. Ofall triangles having the same base and equal areas, that which is 
isosceles has the least perimeter; and of all triangles having the same base 
and equal perimeters, that which is isosceles has the greatest area. 

Ex. 42. Of all triangles having equal areas, that which is equilateral has 
the least perimeter; and of all triangles having equal perimeters, that which 
is equilateral has the greatest area. 

Ex. 43. Of all plane figures having equal perimeters, the circle has the 
greatest area; and of all plane figures having equal areas, the circle has the 
least perimeter. 

Ex. 44. A polygon which can be inscribed in a circle has a greater area than 
any other one constructed with the same sides, 

Ex. 40, A regular polygon has a greater area than any other polygon of the 
same perimeter and number of sides; and it has a less perimeter than any 
other of the same area and number of sides. 

Ex. 46. The lines joining the middle points of the opposite sides of a 
quadrilateral meet and bisect each other, whether the quadrilateral be all 
in the same plane or not; and Ex. 5 is true of any quadrilateral, plane or 
not. 


APPENDIX. 4197 


Ex. 47. In any quadrilateral, plane or not, the sum of the squares of the 
diagonals is equal to twice the sum of the squares of the lines joining the 
middle points of the opposite sides. 

Ex. 48. In any polyedron, the number of its edges increased by two, is 
equal to the number of its vertices increased by the number of its faces. 

Ex. 49. The sum of all the angles of the faces of any polyedron is equal to 
four right angles taken as many times as the polyedron has vertices less two. 

Ex. 50. A perpendicular dropped on any plane from the medial point of any 
triangle (see Ex. 3), is the arithmetical mean of the perpendiculars dropped 
from the angles of the triangle on the same plane. The perpendiculars must 
be taken with opposite signs if they lie on opposite sides of the plane. 

Ex. 51. If the vertices of a tetraedron be connected with the medial points 
of the opposite faces, the four connecting lines will meet at the same point; 
and this point will be three times as far from the vertex as from the medial 
point of the opposite face. 

This point is called the medial point of the tetraedron. 

Ex. 52. A perpendicular dropped on any plane from the medial point of a 
tetraedron is the arithmetical mean of the perpendiculars dropped on the 
same plane from the vertices of the tetraedron. The perpendiculars must be 
taken with opposite signs if they lie on opposite sides of the plane. 

Ex. -53. In any tetraedron, the lines joining the middle points of the oppo- 
site edges meet and bisect each other at the same point. 

Ex. 54. Any plane passing through one of these lines divides the tetraedron 
into two solids equal in volume. 

Ex. 55. Through any four points not in the same plane, one, and only one, 
sphere can be passed. 

Ex. 56. A sphere may be inscribed in any given tetraedron. 


AoPs PoGiIN= Ds HEx@iaos 


THE following problems are also appended, to serve, like the propositions, 
as exercises for the ingenuity of the student. 

The solution of a geometrical problem must be made by geometrical con- 
struction, not by any process of computation combined with the use of scales 
of equal parts or protractors for the measurement of angles. 

The best way to solve a problem will generally be to regard it as solved, 
and draw a figure representing approximately the construction at which we 
wish to arrive ; then, by the help of previous propositions or solved problems, 
to consider the relations between the data and that construction. 

For example: let it be required to connect the sides of an angle by a line 
drawn through a given point lying in the plane of that angle and between its 
sides, in such a way that the line shall be bisected at the point. ‘ 

Let the angle be ACB and the point D. 

Draw EDF, representing the line which we wish to 
draw. We have then a triangle CEF, in which one 
side is bisected. This suggests Prop. 40, and we see 
that a line DG drawn parallel to BC will bisect CE. 
This readily suggests the following construction. 
Draw through D a line parallel to BC, cutting AC in 
G. Lay off GE=CG. Draw the line ED, cutting 
BC in F;; it will be the line required. 

It will be noticed that for this construction we must 
have previously had a construction for drawing a line 
parallel to a given one, through a given point. There is then an order to be 
observed in the solution of problems, as well as in the demonstration of 
propositions. 

An important class of problems is the construction of Joc?, as they are called. 

A locus (plural loci, as above) is a collection of points, usually forming a 
continuous line or surface, in which each point satisfies the same imposed con- 
dition or conditions. 

A circle, for instance, is a locus, the points being subject to the conditions 
that they must lie in the same plane, and be all at the same distance from 
some point in that plane. 

Most problems regarding loci require the aid of what is called analytical 
geometry; some however can be solved with the means now at the student’s 





command. 
For instance: let it be proposed to find the locus of all the points in a plane 
whose distances from two fixed points in that plane are in a given ratio. 


APPENDIX. 129 


Let the points be A and B. , Two points of the required locus are evidently 
on the line AB, one between A and B, the other beyond either A or B. Let 
these be represented by C and D. Let E be 
some other point of the locus. We shall have 
a = con = oe Hence, by Prop. 51, EC bi- * 
sects AEB; and by a method of proof precisely 
similar to that of that proposition, we can show 4 ¢.B 
that ED bisects the angle formed by ED with 
the prolongation of AE; hence EC and ED are perpendicular to each other. 

Hence (219) if a circle be described around a point half way between C 
with a radius equal to 3CD, all its points will satisfy the required conditions; 
and it can easily be shown that lines drawn from C and D to any other point 
in the plane will not be perpendicular to each other, and hence, will not satisfy 
the conditions, 

The points C and D, and the point half way between them, must be found 
by a separate construction. 

PROBLEM 1. To bisect a given straight line. 

Pros. 2. At a given point of a straight line, to erect a perpendicular to it. 

Pros. 3. From a given point outside of a straight line, to drop a perpen- 
dicular on it. 


Pros. 4, Ata given point on a line, to construct a given angle. 

Props. 5. Through a given point outside of a line, to draw a parallel to it. 

PROB. 6. To bisect a given angle. 

Pros. 7. Two angles of a triangle being given, to construct the third. 

Pros. 8. Two angles of a triangle and a side being given, to construct the 
triangle. 


Pros. 9. Two sides of a triangle and the included angle being given, to 
construct the triangle. 

Pros, 10. Two sides of a triangle and the angle opposite one of them being 
given, to construct the triangle. 

Prop. 11. The three sides of a triangle being given, to construct the tri- 
angle. 

Pros. 12. To divide a straight line into parts proportional to given lines. 

Pros. 13. To construct a fourth proportional to three given lines. 

Pros. 14. To construct a mean proportional between two given lines, 

Pros. 15. To divide a line into two such parts that the greater part shall 
be a mean proportional between the whole line and the lesser part. (As in 
Prop. 77.) 

Pros. 16. To fix a point on a line and also one on its production at either 
end, such that the ratio of their distances from one end to their distances 
from the other shall have any given value, the same for both. 

A line thus divided is said to be divided harmonically. 

Pros. 17. On a given line as side, to construct a polygon similar to a given 
polygon. 

Pros. 18. To trisect a given straight line. (See Ex. 3.) 

Pros. 19. In a given straight line, to find a point equally distant from two 
given points ate of the line. 


130 APPENDIX. 


Pros. 20. From two given points outside of a given straight line, to 
draw two straight lines meeting in the given one and making equal angles 
with it. 

Pros, 21. Two lines being given, the intersection of which is so remote 
that the lines cannot be produced to it, to draw a line which will pass through 
their intersection and bisect the angle formed by them; also to draw one 
which will pass through their intersection and any other given point lying 
in their plane. 

PROB. 22. Given two lines in the same plane, to draw through a point 
outside, a line making twice as great an angle with one as with the other. 

Pros. 28. To draw a line through a point outside of another line, so as to 
make a given angle with that line. 

Pros, 24. Given for a triangle one angle, an adjacent side, and the sum or 
difference of the other two sides, to construct the triangle. 

Pros. 25. Given for a triangle, one angle, an opposite side and the sum or 
difference of the other two sides, to construct the triangle. 

Pros. 26. Given for a triangle, one side, the difference of the angles at its 
extremities, and the sum of the other two sides, to construct the triangle. 

Pros. 27. Given for a triangle, the three medial lines, to construct the 
triangle. 

Pros. 28. Given for a triangle, one angle, the side opposite to it, and the 
ratio of the other two sides, to construct the triangle. 

PRos. 29, Given for a triangle, the angles and perimeter, to construct the 
triangle. 

Pros. 30. Given, in a triangle, the angles and area, to construct the tri- 
angle. 

Pros. 31. Given, in a triangle, two sides and the length of the line bisect- 
ing the included angle and terminating in the opposite side, to construct the 
triangle. 

PROB. 32. Given, in a right-angled triangle, the differences of the sides, to 
construct the triangle. 

Pros. 33. Given, the distances to three of the vertices of a square from a 
point lying in the plane of a square, to construct the square. 

Pros. 34. To draw through a point lying in the plane of two lines inter- 
secting each other, a line such that the parts into which it is divided at the 
‘point shall be in a given ratio. 

Pros, 35. To bisect the area of a given triangle by a line drawn through a 
given point in one of its sides. 

Pros. 386. To divide the area of a given triangle into any number of parts 
having given ratios to each other, by lines parallel to one of its sides. 

Pros. 87. To construct a rectangle having the same perimeter as a given 
square and half the area. 

Pros. 38. Given, the difference of the sides of two squares, one of which 
contains twice the area of the other, to construct the squares. 

Pros. 39. To divide a given angle into two parts, such that the two per- 
‘ pendiculars from the same point of the dividing line on the sides of the angle 
shall be in a given ratio. : 

Pros, 40. To find a point within a given triangle, such that the lines drawn 


APPENDIX. 131 


from it to the vertices of the triangle shall divide the area of the triangle into 
three equal parts. 

Pros. 41. To find a point within a given triangle, such that the lines drawn 
from it to the middle points of the sides of the triangle shall divide the area 
of the triangle into three equal parts. 

Pros. 42. To find a point on a straight line the sum of the distances from 
which to a number of given points on the line,is zero; the distances being 
taken with opposite algebraic signs if they lie on opposite sides of the required 
point. 

Pros, 43. To find a point on a straight line, the sum of the squares of the 
distances from which to a number of given points on the line is a minimum. 

Pros. 44. To find the point on a plane, or in space, the sum of the squares 
of the distances from which to a number of given points on the plane or in 
space is a minimum. 

PROB. 45. To inscribe a square in a given triangle. 

Pros, 46. To inscribe a parallelogram in a given triangle, so that it shall 
have the greatest possible area. 

Pros. 47. To find a point on a given line, the sum of the distances of which 
from two points outside shall be a minimum. 

Pros. 48. To construct a triangle equal to a given polygon. 

Pros, 49. To construct a square equal to a given triangle. 

Pros. 50. To construct a polygon similar to one polygon and equal in area 
to another. 

Pros. 51. To draw a tangent to a circle, from a point on the circle or 
outside. 

Pros. 52. To draw a common tangent to two circles. 

Pros. 53. To inscribe a circle in a given triangle. 

Pros. 54, To circumscribe a circle about a given triangle. 

Pros. 55. On a given line as chord, to describe an are which shall contain 
a given inscribed angle. 

Pros. 56. (Three point problem.) Three points being given, to find a point 
in their plane, such that the lines drawn from it to the three points shall make 
given angles with each other. 

Pros. 57. Three points being given, to find a point in their plane, such that 
the lines drawn from it to the three points shall have given ratios to each 
other. 

Pros. 58. Three points being given, to draw a line in their plane through 
one of them, such that the perpendiculars on it from the other two shall be in 
a given ratio. 

Pros. 59. Through a given point in the plane of two parallels, to draw a 
line cutting the parallels so that the distance of its intersection with one of 
them from a given point on that one shall be in a given ratio to the distance 
of its intersection with the other from a given point on that other. 

Pros. 60. To draw three circles, each of which shall be tangent to one of 
the sides of a triangle, and the other two sides produced. 

These circles are called escribed circles. The sides of the triangle formed by 
their centres will pass through the vertices of the original triangle. 

Pros. 61. To construct a triangle, given the centres of its escribed circles, 


182 APPENDIX. 


Pros. 62, Around three given points as centres, to draw three circles tan- 
gent to each other. 

Pros. 63. Given an angle and a point in its plane, to draw through the 
point a line which shall make, with the sides of the angle, a triangle of a 
given perimeter. 

Pros. 64. The same being given, to draw a line which shall make a tri- 
angle of a given area. 

Pros. 65. To find a point ina given are of a pivale, such that the chords 
drawn from it to the ends of the arc shall be in a given ratio. 

Pros. 66. Describe a circle passing through two given points and tangent 
to a given line, the points and the line being in the same plane. 

Pros. 67. Describe a circle passing through two given points and tangent 
to a given circle, the points and the circle being in the same plane. 

Pros. 68. Describe a circle passing through a given point and tangent to 
two given lines, the point and the lines being in the same plane. 

Pros. 69. Describe a circle passing through a given point and tangent to 
two given circles, the point and the circles being in the same plane. 

Pros. 70. Describe a circle tangent to three given circles in the same plane. 

Pros. 71. Draw a tangent to a given circle, such that the part of it inter- 
cepted between two given produced radii of the circle shall be equal to given 
line. 

Pros..72. Draw a tangent to a given circle, such that the part of it inter- 
cepted between two given tangents to the same circle, shall be equal to a 
given line. 

Pros. 73. In a given circle, to place two chords of given length, and in- 
clined at a given angle. 

‘ Pros. 74. From a given point in one of the sides of a triangle, to draw three 
lines trisecting the area of the triangle. 

Pros. 75. Given one side of a triangle, and the radii of the inscribed and 
circumscribed circles, to construct the triangle. 

Pros. 76. Through a given point, with a given radius, to describe a circle 
bisecting a given circle. 

Pros. 77. With a given radius to describe a circle bisecting two given circles. 

Pros. 78. Bisect the area of a trapezoid by a line drawn parallel to its 
parallel sides. 

Pros. 79. Bisect the area of a triangle by the shortest possible line. 

Pros. 80. Bisect the area of a parallelogram by a line such that if it be 
produced to meet the sides which it does not meet unless they are produced, 
the areas of the external triangles will have a given ratio to that of the paral- 
lelogram. 

Pros. 81. Divide the area of a given circle into parts, having any required 
ratios, by concentric circles. 

Pros. 82. In an equilateral triangle, inscribe three equal circles tangent to 
each other and to the sides of the triangle. 

Pros. 83. In a circle, inscribe three equal circles tangent to each other and 
to the original circle. 

Pros. 84. To construct a triangle similar to a given triangle, and having 
its vertices on three given parallel lines, 


APPENDIX. 133 


Pros. 85. The same, substituting ‘‘ concentric circles’’ for ‘‘ parallel lines.” 

Pros. 86. Find the locus of all the points in a plane, the difference of the 
squares of the distances of which from two given points is equal to the square 
of a given line. 

Pros. 87. Find the locus of all the points in a plane, the sum of the dis- 
tances of which from two given lines is equal to a given line. 

Pros. 88. Find the same, substituting ‘‘ sum” for ‘‘ difference.” 

Pros, 89. Find the same, counting the distances as of opposite signs when 
they fall on opposite sides of the lines. 

Pros. 90. Given a circle and a fixed point outside, in the plane of the cir- 
cle; find the locus of the point from which the line drawn to the fixed point 
and the tangent to the circle shall be equal. 

Pros, 91. Find the locus of the middle point of a line of given length con- 
necting two lines perpendicular to each other. 

Pros. 92. From any point, lines are drawn toa given line, and divided 
internally or externally, so that the product of the distances from the point 
and line to the point of division is constant; find the locus of the point of 
division. 

Pros. 93. The same, substituting a given circle for the given line. 

Pros. 94. From a point lines are drawn to a given circle and divided in a 
given ratio; find the locus of the point of division. 

Pros. 95. Upon a given side a triangle is constructed, having a given angle 
- opposite to this side; find the locus of the point mentioned in Ex. 2. 

Pros. 96. In the same conditions, find the loci of the centres of the in- 
scribed and escribed circles. ; 

Pros. 97. Two circles being drawn in the same plane; find the locus of 
the point in that plane from which these circles would appear equal. 

Pros. 98. Find the locus of a point such that the sum of the squares of its 
distance from any number of points in a plane is equal to the square of a 
given line. 

Prog. 99. The same, substituting ‘‘ space” for ‘‘a plane.” 


Note.—Other problems in space are not given, the subject more properly 
belonging to Descriptive Geometry. 


APPENDIX 3G: 


Note to page 23. 


THE proposition that ‘‘ from any point of space a line may be drawn in the 
same direction with any given line”? has been introduced (81) as an assump- 
tion, the truth of which seems to be self-evident. 

It may be well to append some remarks upon this assumption, which is the 
one which has been chosen (as some one must be chosen), as the basis of the 
theory of parallels adopted in this work. 

In the first place, then, it may be shown that from any point in space a line 
may be drawn making an indefinitely small angle with any given line. 

For this, we may first establish the following proposition : 

In any triangle, any angle is a smaller fraction of two right angles than the 
side opposite tt ts of the longest side. 

If the side opposite it is as long as any, the proposition is self-evident, for 
it is merely saying that the angle is 
less than two right angles; we will 
therefore suppose, in the triangle 
ABC, that AC is the longest side, or” 
at least as long as BC, and longer than 
AB. 

Drop AD perpendicular to AB, or 
to AB prolonged (as in the figure), if 
necessary. It will not fall at C, or on the prolongation of BC beyond C, for 
ACB is acute (41, 38), since AB < AC. 

Lay off on AD prolonged, DA’ = AD, and draw CA’. ACA’ is of course an 
isosceles triangle. Construct A’CA” equal to it in the same plane, and con- 
tinue constructing these triangles till the angle ACA” (counted from CA 
around to the right), is equal to, or greater than 2R. 

Prolong A@C till it meets the line AA’, which of course it will do at a 
distance OF less than AC. 

The broken line AMA®™—-V ,, . A” A'F is longer than the straight line 
A"F; hence AVA®-Y . . . A’ A’A is longer than AF + FA, and still more is 
it longer than A”C + CA, since CA < CF + FA. Or, it is longer than 2CA. 

But AD = Len eens be Bs 

2n 


greater than i And the angle ACB = 





; hence itis greater, and still more is AB 


ACA®—) 2k 
2(n Tyee = Snel), 

But 2 must be at least 2, since ACB. is, as previously shown, acute; hence 
2(n — 1) is at least as great asm; hence the angle ACB, is, as was to be 
proved, a smaller fraction of 2R than AB is of CA. 





= APPENDIX. 135 


It follows from this, that, if we drop from any point outside of a line a per- 
pendicular on that line, and lay off a distance on that line from the foot of the 
perpendicular, equal to m times that perpendicular. (m being any number 
whatever), and connect the end of this distance with the point outside of the 
line, the angle which this connecting line makes with the original one will be 


less than ae ie 
; m 
Consequently, if we wish to make the angle less than any value whatever, 
2k . 
say than 7, we have only to make m = ee The angle may then be made 


indefinitely small; which was what we wished to show. 
Now, if we rotate the connecting line around the point outside, away from 
the other line, starting from the position which it has when the angle which 
it forms with the other is indefinitely small, the rotation being made in the 
plane which it forms with the other one, it is plain that when the rotation is 
equal to 2R minus twice the angle which it now forms with the perpendicular, 
it will meet the other line on the other side of the perpendicular, making an 
equal indefinitely small angle with it, and will, in accordance with generally 
received principles (4), fail to meet it on the side on which it now does. There 
must then be some definite position in this rotation which is the limit between 
those in which it meets it on the side on which it now does and those to 
_which its final position belongs, in which it fails to do so. At this limit, it 
would seem that the angle which it makes with the other line absolutely dis- 
appears, since it has previously been indefinitely small, and that its direction, 
at the infinitely removed point at which it finally quits the other line, and con- 
sequently also at any other point, is the same as that of that other line. 
The above explanation may, perhaps, help to justify our assumption. It 
has not been intended as a rigorous demonstration of it; it is not probable 
that the obscurity will ever be completely removed from the theory of parallels. 





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